Binomial Expansion Part2

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Binomial Coefficient Revisited

Let's look at the first few binomial coefficients

  • \binom{n}{0} =\frac{n!}{\left(n-0\right)!0!} =\frac{n!}{n!1} =1\,
  • \binom{n}{1} =\frac{n!}{\left(n-1\right)!1!} =\frac{n\left(n-1\right)!}{\left(n-1\right)!} =n\,
  • \binom{n}{2} =\frac{n!}{\left(n-2\right)!2!} =\frac{n\left(n-1\right)\left(n-2\right)!}{\left(n-2\right)!2!} =\frac{n\left(n-1\right)}{2!}\,
  • \binom{n}{3} =\frac{n!}{\left(n-3\right)!3!} =\frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)!}{\left(n-3\right)!3!} =\frac{n\left(n-1\right)\left(n-2\right)}{3!}
  • \binom{n}{4} Can you guess the value from the pattern of the first few? \frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}
  • And the next one will be \frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)}{5!}

Expansion for Non Positive integer n

{\left(1+x\right)}^{n} =1+nx+\frac{n\left(n-1\right)}{2!}{x}^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^{3}+\ldots

  • Valid for \left|x \right| <1

At a glance, it seems to be the same formula we have already learnt for positive integer n\,, just with the binomial coefficients written in the other form (refer above), but there are fundamental differences between the two.

Important Differences

  • Non positive integer n\, here includes
    • negative integers, fractions, negative fractions
  • It is an infinite series (never ending), and we MUST write +\ldots at the end. Normally we would be told which term we need to expand until.
    • Compare with positive integers n\,
      • {\left(1+x\right)}^{3} will end at {x}^{3} \,
      • {\left(1+x\right)}^{10} will end at {x}^{10} \, (but we might be asked to expand until certain terms only)
      • {\left(1+x\right)}^{-2} will have a never-ending expansion (but we WILL be told to expand till where)

  • We MUST use {\left(1+x\right)}^{n} form, there is no {\left(a+b\right)}^{n} form for non positive integer n\,
    • For positive integers we can use either.
  • Expansion is only valid for \left|x \right| <1
    • For positive integers n\, it is valid for all values.
  • For non positive integers, \binom {n}{r} does not exist (your calculator will give an error), thus we MUST use the other representation of the coefficients.
    • However, can we use that alternative representation for positive integer? Yes.

Of Apples & Oranges & Bananas

Actually, there are only two variables here, x,n\, (the 1\, is fixed). Lets see how we are going to use the formula.

  • {\left(1+x\right)}^{n}=1+nx+\frac{n\left(n-1\right)}{2!}{x}^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^{3}+\ldots
  • {\left(1+{\color{Red}\heartsuit}\right)}^{{\color{Blue}\triangle}} =1+{\color{Blue}\triangle}{\color{Red}\heartsuit}+\frac{{\color{Blue}\triangle}\left({\color{Blue}\triangle}-1\right)}{2!}{{\color{Red}\heartsuit}}^{2}+\frac{{\color{Blue}\triangle}\left({\color{Blue}\triangle}-1\right)\left({\color{Blue}\triangle}-2\right)}{3!}{{\color{Red}\heartsuit}}^{3}+\ldots
  • So we always start with 1\,
  • Followed by "nx\,", which is the power times whatever we use as "x\,"
  • The following terms,
    • the "x\," we just keep increasing the power
    • the "n\,", follows a set pattern, each term having "more" compared to the previous n \, \frac{{\color{Red}n}\left(n-1\right)}{{\color{Blue}2!}} \frac{{\color{Red}n\left(n-1\right)}\left(n-2\right)}{{\color{Blue}3!}} \frac{{\color{Red}n\left(n-1\right)\left(n-2\right)}\left(n-3\right)}{{\color{Blue}4!}}

Calculating Coefficients

Lets say we have n=-1\,,

  • the coefficient \frac{n\left(n-1\right)\left(n-2\right)}{3!} will become \frac{{\color{Red}-1}\left({\color{Red}-1}-1\right)\left({\color{Red}-1}-2\right)}{3!}, which, though correct, is messy to calculate, not to mention creates a high risk of careless mistake
  • a better way to write it is to first understand n\left(n-1\right)\left(n-2\right) as series of steps
    • take n\,
    • minus one and multiply it
    • minus one again and multiply it
  • Thus, we will write
    • {\color{Red}\left(-1\right)} then minus 1\, which will give us -2\,
    • \left(-1\right){\color{Red}\left(-2\right)} then take \left(-2\right) and minus 1\,again which will give us -3\,
    • \left(-1\right)\left(-2\right){\color{Red}\left(-3\right)}
    • Thus \frac{\left(-1\right)\left(-2\right)\left(-3\right)}{3!}
    • In fact, note that we would have already written \frac{\left(-1\right)\left(-2\right)}{2!} earlier, thus we actually do one extra step only

Note : Again, the pattern of the signs is either

  1. all positive
  2. all negative (except first term)
  3. positive and negative alternately

Examples

Expand the following in ascending powers of x\, until the term in {x}^{3}\,

  • {\left(1+x\right)}^{-1}
    • Let's write it out carefully
      • 1\,
      • 1+\left(-1\right)x DON'T write -x \, yet.
      • 1+{\color{Red}\left(-1\right)}x+\frac{{\color{Red}\left(-1\right)}\left(-2\right)}{2!}{x}^{2}
      • 1+\left(-1\right)x+\frac{{\color{Blue}\left(-1\right)\left(-2\right)}}{2!}{x}^{2}+\frac{{\color{Blue}\left(-1\right)\left(-2\right)}\left(-3\right)}{3!}{x}^{3} Notice how we just keep on "expanding" the coefficients from previous one.
      • Finished? Sure? Must put +\ldots
    • =1+\left(-1\right)x+\frac{\left(-1\right)\left(-2\right)}{2!}{x}^{2}+\frac{\left(-1\right)\left(-2\right)\left(-3\right)}{3!}{x}^{3}+\ldots
    • =1-x+{x}^{2}-{x}^{3}+\ldots


  • {\left(1+x\right)}^{\frac{1}{2}}
    • calculate the fractions carefully
      • 1\,
      • 1+\left(\frac{1}{2}\right)x
      • 1+{\color{Red}\left(\frac{1}{2}\right)}x+\frac{{\color{Red}\left(\frac{1}{2}\right)}\left(-\frac{1}{2}\right)}{2!}{x}^{2}
      • 1+\left(\frac{1}{2}\right)x+\frac{{\color{Blue}\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}}{2!}{x}^{2}+\frac{{\color{Blue}\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}\left(-\frac{3}{2}\right)}{3!}{x}^{3}
    • =1+\left(\frac{1}{2}\right)x+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{x}^{2}+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}{x}^{3}+\ldots To calculate\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!} using calculator, rather than typing the whole thing (waste time and easy to press wrongly), we might as well take 3\, (product of numerators) divide by 8\, (product of denominators) and then divide by 3!\,, and then determine the sign.
    • =1+\frac{1}{2}x-\frac{1}{8}{x}^{2}+\frac{1}{16}{x}^{3}+\ldotsNote: Write \frac{1}{2}x instead of \frac{x}{2}


Try these on your own :

  • {\left(1-x\right)}^{-\frac{1}{2}} Answer : =1+\frac{1}{2}x+\frac{3}{8}{x}^{2}+\frac{5}{16}{x}^{3}+\ldots
    • If you got it wrong, it's probably because while paying attention to the n\, your forgot about the "apple".
    • =1+\left(-\frac{1}{2}\right){\color{Red}\left(-x \right)}+\frac{\left(-\frac{1}{2}\right)\left( -\frac{3}{2}\right)}{2!}{{\color{Red}\left(-x \right)}}^{2} +\frac{\left(-\frac{1}{2}\right)\left( -\frac{3}{2}\right)\left( -\frac{5}{2}\right)}{3!}{{\color{Red}\left(-x \right)}}^{3}+\ldots


  • {\left(1+2x\right)}^{-2} Answer : =1-4x+12{x}^{2}-32{x}^{3}+\ldots
    • =1+\left(-2\right){\color{Red}\left(2x\right)}+\frac{\left(-2\right)\left(-3\right)}{2!}{{\color{Red}\left(2x\right)}}^{2} +\frac{\left(-2\right)\left(-3\right)\left(-4\right)}{3!}{{\color{Red}\left(2x\right)}}^{3}+\ldots

Things To Do Before Expanding

Usually, there are a few steps that need to be done before we can even start using that formula. Do note that if there are mistakes in these early steps, everything else will be wrong and just a waste of time.

Determining value of n

Make sure you don't make a mistake in this step

  • \frac{1}{1+x} ={\left(1+x \right)}^{-1}
  • \frac{3}{1+x} =3{\left(1+x \right)}^{-1} Note: There is no reason to disturb the 3\,, just leave it as it is.
  • \frac{1}{3\left(1+x \right)} =\frac{1}{3}{\left(1+x \right)}^{-1} Note: There is, again, no reason to disturb the 3\,
  • \sqrt{1+x} ={\left(1+x \right)}^{\frac{1}{2}}
  • \sqrt[3]{1+x} ={\left(1+x \right)}^{\frac{1}{3}} Note: and NOT \frac{2}{3}
  • \frac{1}{\sqrt{1+x}} ={\left(1+x \right)}^{-\frac{1}{2}}

Changing Into (1+x)^n

In the formula

  • it MUST be a 1\,
  • "x\," can be anything

So what do we do if the question is NOT a 1\,?

Suppose we have \left({\color{Red}2}+4x \right), we can change the 2\, to a 1\, by factorizing the 2\,.

  • \left(2+4x \right) 2\left(1+2x \right)

What if we have \left(x+2\right) ?

  • \left(x+2\right) =\left(2+x\right) \because a+b=b+a
    • =2\left(1+\frac{x}{2}\right) Now, the \frac{x}{2} will look troublesome, but remember we DON'T have a choice, we MUST make the first term to be 1\,

But the above is without any power. How will n\, affect it?

Remember that {\left(ab \right)}^{n} ={a}^{n}{b}^{n}\,

In other words, when we want to bring the a\, out, we must apply the power n\, to it too.

  • {\left(2+4x\right)}^{n}  ={\left[2\left(1+2x\right)\right]}^{n} ={2}^{n}{\left(1+2x\right)}^{n}


Try these

  • {\left(2+x\right)}^{-1} 
={\left[2\left(1+\frac{x}{2}\right)\right]}^{-1} 
=\frac{1}{2}{\left(1+\frac{x}{2}\right)}^{-1}
  • \sqrt{x+4} 
={\left(4+x \right)}^{\frac{1}{2}} 
={\left[4\left(1+\frac{x}{4}\right)\right]}^{\frac{1}{2}}  =2{\left(1+\frac{x}{4}\right)}^{\frac{1}{2}}
  • {\left(2x-1\right)}^{-3} 
={\left[-\left(1-2x\right)\right]}^{-3} 
=-{\left(1-2x\right)}^{-3} Note: \because {\left(-1 \right)}^{-3}=-1
  • \sqrt[3]{3x+27} 
={\left(27+3x \right)}^{\frac{1}{3}} 
={\left[27\left(1+\frac{1}{9}x\right)\right]}^{\frac{1}{3}}  =3{\left(1+\frac{1}{9}x\right)}^{\frac{1}{3}}

Multiple Expansions

Keep these in mind

  • its always easier to +/-\, than \times/\div
    • Thus, if applicable, we would expand the partial fraction form
  • it is very difficult to divide polynomial
    • Multiplying, while a lot of work, is still doable. But how do we change divide into times? \frac{a}{b}=a\left(?\right) \frac{a}{b}=a{b}^{-1}


  • \frac{1}{\left(1+x\right)}-\frac{1}{\left(1+2x\right)} ={\left(1+x\right)}^{-1}-{\left(1+2x\right)}^{-1}
  • \frac{1+x}{1-x} =\left(1+x\right){\left(1-x\right)}^{-1}
  • \frac{{\left(1+x \right)}^{2}}{{\left(1+3x \right)}^{3}} ={\left(1+x \right)}^{2}{\left(1+3x \right)}^{-3}
    • We could have expanded top and bottom, but we will end up with a polynomial divide polynomial.
  • \sqrt{\frac{1+x}{1-x}} ={\left(1+x\right)}^{\frac{1}{2}}{\left(1-x\right)}^{-\frac{1}{2}}
  • \sqrt[3]{\frac{1+x}{{\left(1+2x \right)}^{2}}}  ={\left(1+x\right)}^{\frac{1}{3}}{\left(1+2x\right)}^{-\frac{2}{3}}

Example

Let's put all these to work:

Expand the following in ascending powers of x\,, neglecting the terms in {x}^{3}\, and above

  • \frac{\sqrt{x+9}}{{\left(1-4x \right)}^{3}}
    • Analyze : There is no point expanding top and bottom. Instead, we must first rearrange it.
    • \frac{\sqrt{x+9}}{{\left(1-4x \right)}^{3}} ={\left(x+9\right)}^{\frac{1}{2}}{\left(1-4x \right)}^{-3}
    • Since both expansion is going to require some work, better we do it separately first
    • {\left(x+9\right)}^{\frac{1}{2}} Analyze : We need to do some work first since it is not 1\,
      • ={\left(9+x\right)}^{\frac{1}{2}} ={\left[9\left(1+\frac{x}{9}\right)\right]}^{\frac{1}{2}} =3{\left(1+\frac{x}{9}\right)}^{\frac{1}{2}} Now we can expand. What about the 3\,? Just leave it outside first.
      • =3\left[1+\left(\frac{1}{2}\right)\left(\frac{x}{9}\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{\left(\frac{x}{9} \right)}^{2}+\dots\right]
      • =3\left(1+\frac{1}{18}x-\frac{1}{648}{x}^{2}+\ldots \right)
      • =3+\frac{1}{6}x-\frac{1}{216}{x}^{2}+\ldots
    • {\left(1-4x \right)}^{-3} Analyze : It's already 1\,, so we just expand
      • =1+\left(-3 \right)\left(-4x \right)+\frac{\left(-3 \right)\left(-4 \right)}{2!}{\left(-4x \right)}^{2}+\ldots
      • =1+12x+96{x}^{2}+\ldots
    • Next step would be multiplying them. Note: DO spend some time checking both expansions before multiplying.
    • \therefore \frac{\sqrt{x+9}}{{\left(1-4x \right)}^{3}} =\left(3+\frac{1}{6}x-\frac{1}{216}{x}^{2}+\ldots\right)\left(1+12x+96{x}^{2}+\ldots\right)
    • =3+36x+288{x}^{2}+\frac{1}{6}x+2x^{2}-\frac{1}{216}{x}^{2}+\ldots
    • =3+\frac{217}{6}x+\frac{62639}{216}{x}^{2}+\ldots
    • Valid for Read the part on validity (below) first and then come back and try
      • \left|\frac{x}{9}\right|<1 and \left|-4x\right|<1
      • \therefore \left|x\right|<9 and \left|x\right|<\frac{1}{4}
      • \therefore \left|x\right|<\frac{1}{4}
      • \therefore -\frac{1}{4}<x<\frac{1}{4}

Comparison

As always, let's take a breather to compare the different things we have learnt thus far.

  • For positive integer n\,
    • {\left(a+b\right)}^{n} ={a}^{n}+\binom{n}{1}{a}^{n-1}b+\binom{n}{2}{a}^{n-2}{b}^{2}+\dots+{b}^{n}
    • {\left(1+x \right)}^{n} =1+\binom{n}{1}x+\binom{n}{2}{x}^{2}+\dots+{x}^{n}
      • =1+nx+\frac{n\left(n-1\right)}{2!}{x}^{2}+\dots+{x}^{n}
  • For non positive integer n\,
    • {\left(1+x\right)}^{n} =1+nx+\frac{n\left(n-1\right)}{2!}{x}^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^{3}+\ldots
    • Valid for \left|x \right| <1
  • If the first term is NOT 1\,
    • For positive integer n\, : Don't need (but can) change it to 1\,
    • For non positive integer n\, : MUST change it to 1\,
  • Coefficients
    • For positive integer n\, : Can use \binom{n}{1},\binom{n}{2} \ldots or n, \frac{n\left(n-1\right)}{2!},\ldots form
    • For non positive integer n\, : MUST use n, \frac{n\left(n-1\right)}{2!},\ldots form
  • Other difference
    • For positive integer n\, : Finite expansion. Valid for all values
    • For non positive integer n\, : Infinite expansion. Valid for \left|x \right| <1

Validity

Let's have a look at the formula again

  • {\left(1+x\right)}^{n} =1+n{\color{Red}x}+\frac{n\left(n-1\right)}{2!}{\color{Red}{x}^{2}}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{\color{Red}{x}^{3}}+\ldots

The expansion is an infinite series, thus as r \to \infty, we must have {x}^{r} \to 0 (so that the series converges)

Note : Otherwise, we will keep on adding more and more terms which is increasingly large, thus {\left(1+x\right)}^{n} \to \infty (which is of course wrong). We can compare this with geometric series, where sum to infinity only exists if \left| r \right|<1, so that as n \to \infty, {r}^{n}\to 0


For {x}^{r} \to 0 as r \to \infty to be true, we must have \left| x \right|<1

Thus, the expansion is only true when \left| x \right|<1

We say the expansion is valid only for \left| x \right|<1 or -1<x<1 \,.

Important

  • The "x\," here refer to WHATEVER we use "x\,"
  • Final answers are normally written in the -1<x<1 \, form, but the modulus form is useful to simplify

Simplifying the Validity Range

  • Simplify the modulus form FIRST before changing to the other from
    • \left|3x\right|<1 \therefore \left|x\right|<\frac{1}{3} \therefore -\frac{1}{3}<x<\frac{1}{3}
    • \left|-\frac{x}{4}\right|<1 \therefore \left|x\right|<4 (modulus takes care of the negative sign already) \therefore -4<x<4
  • Since \left|x\right| is always positive, we CAN multiply it across
    • \left|\frac{1}{x}\right|<1 \therefore 1<\left|x\right| \therefore x<-1 or x>1\,
    • \left|-\frac{3}{2x}\right|<1 \therefore \frac{3}{2}<\left|x\right| \therefore x<-\frac{3}{2} or x>\frac{3}{2}\,
  • {x}^{2}\, is always positive (\geq 0 actually)
    • \left|{x}^{2}\right|<1 \therefore {x}^{2}<1 (modulus sign is not needed anymore)
    • \left|x+{x}^{2}\right|<1 We can't take away the modulus for this one, however, and this would need to be solved slowly using normal inequalities method.
  • Combination
    • \left|x \right|<3 and \left|x \right|<2 \therefore \left|x \right|<2
    • \left|x \right|>3 and \left|x \right|>2 \therefore \left|x \right|>3
    • \left|\frac{x}{2}\right|<1 and \left|3x \right|<1 \therefore \left|x\right|<2 and \left|x \right|<\frac{1}{3} \therefore \left|x \right|<\frac{1}{3} \therefore -\frac{1}{3}<x<\frac{1}{3}

Identifying the x

The "x\," that we are going to put into \left|x\right| is the same whatever "x\," we use in the expansion. In other words

  • {\left(1+{\color{Red}\heartsuit}\right)}^{n} =1+n{\color{Red}\heartsuit}+\frac{n\left(n-1\right)}{2!}{{\color{Red}\heartsuit}}^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{{\color{Red}\heartsuit}}^{3}+\ldots
  • Valid for \left|{\color{Red}\heartsuit}\right|<1\,

Examples

  • {\left(1+x \right)}^{-1}
    • Valid for \left|x \right|<1
  • {\left(1-3x \right)}^{\frac{1}{2}} {\left(1{\color{Red}-3x} \right)}^{\frac{1}{2}}
    • Valid for \left|-3x \right|<1
  • {\left(1+{x}^{2} \right)}^{-1}
    • Valid for \left|{x}^{2} \right|<1
  • {\left(2+x \right)}^{-3} =\frac{1}{8}{\left( 1+{\color{Red}\frac{x}{2}}\right)}^{-3} We must see from this step, right before we use the formula (rather than the earliest form)
    • Valid for \left|\frac{x}{2} \right|<1
  • {\left(1+x \right)}^{-1}-{\left(1+2x \right)}^{-1} {\left(1+{\color{Red}x} \right)}^{-1}-{\left(1+{\color{Blue}2x} \right)}^{-1}
    • Valid for \left|x \right|<1 and \left|2x \right|<1
  • {\left(1+x \right)}^{\frac{1}{2}}{\left(1+2x \right)}^{-\frac{1}{2}}
    • Valid for \left|x \right|<1 and \left|2x \right|<1
  • \left(1+x \right){\left(1+2x \right)}^{-\frac{1}{2}} \left(1+x \right){\left(1+{\color{Red}2x}\right)}^{-\frac{1}{2}} The first bracket DON'T need since we did not expand it
    • Valid for \left|2x \right|<1
  • {\left(1+2x \right)}^{3}{\left(1+x \right)}^{-2} {\left(1+2x \right)}^{3}{\left(1+{\color{Red}x} \right)}^{-2} The first bracket DON'T need since it's expansion for positive integer n\, (which is valid for all)
    • Valid for \left|x \right|<1
  • \frac{x+2}{x+3} =\left(2+x\right){\left(3+x\right)}^{-1}=\left(2+x\right)\left[{\frac{1}{3}\left(1+{\color{Red}\frac{x}{3}}\right)}^{-1}\right]
    • Valid for \left|\frac{x}{3} \right|<1

Stating the Validity

What form we state it depends on the form of the question

  • Question : State the range of x\, where/such that the expansion is valid
    • Answer : Valid for -1<x<1\,
  • Question :State the set of values of x\, where/such that the expansion is valid
    • Answer : Valid for \left\{ x: -1<x<1 \right\} (without set notation, marks won't be given)
  • Question :Find set S\, where the expansion is valid when x \in S
    • Answer : Valid for S=\left\{ x: -1<x<1 \right\} (just a fancy way to ask you for the set)

Normally, the question will state it clearly if you are required to state the validity range. If not asked to do so, you need not state it, but you might want to write it out for your own reference when doing approximation later.

Exercise 5

Notes

  • Certainly a lot of things here to be taken care of. Make sure you are very familiar with all the above before attempting this.


1) Expand the following in ascending powers of x\, until the terms in {x}^{3}\, and above.

  • a) {\left(1-x\right)}^{-1} =1+x+{x}^{2}+{x}^{3}+\ldots
  • 
\begin{align}
{\left(1-x\right)}^{-1}
&=1+\left(-1\right)\left(-x\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(-x\right)}^{2}+\frac{\left(-1\right)\left(-2\right)\left(-3\right)}{3!}{\left(-x\right)}^{3}+\ldots\\ 
& =1+x+{x}^{2}+{x}^{3}+\ldots 
\end{align}
  • b) {\left(1+3x\right)}^{-\frac{1}{2}} 1-\frac{3}{2}x+\frac{27}{8}{x}^{2}-\frac{135}{16}{x}^{3}+\ldots
  • 
\begin{align}
{\left(1+3x\right)}^{-\frac{1}{2}}
&=1+\left(-\frac{1}{2}\right)\left(3x\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}{\left(3x\right)}^{2}+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}{\left(3x\right)}^{3}+\ldots\\ 
& =1-\frac{3}{2}x+\frac{27}{8}{x}^{2}-\frac{135}{16}{x}^{3}+\ldots 
\end{align}

2) Expand the following in ascending powers of x\,, neglecting the terms in {x}^{3}\, and above. State the set of values of x\, where the expansion is valid.

  • a) \frac{3}{2\left(x+4 \right)} \frac{3}{8}-\frac{3}{32}x+\frac{3}{128}{x}^{2}+\ldots\;\mbox{Valid for }\left\{x:-4<x<4\right\}
    • 
\begin{align}
\frac{3}{2\left(x+4 \right)}
&=\frac{3}{2}{\left(4+x \right)}^{-1} \\
&=\frac{3}{8}{\left(1+\frac{x}{4}\right)}^{-1} \\
&=\frac{3}{8}\left[1+\left(-1\right)\left( \frac{x}{4}\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(\frac{x}{4}\right)}^{2}+\ldots\right] \\
&=\frac{3}{8}\left(1-\frac{x}{4}+\frac{{x}^{2}}{16}+\ldots \right) \\
&=\frac{3}{8}-\frac{3}{32}x+\frac{3}{128}{x}^{2}+\ldots \\
\end{align}
    • \begin{align}
& \mbox{Valid for }\left| \frac{x}{4}\right|<1, \therefore \left|x \right|<4 \\
& \therefore \mbox{Valid for }\left\{x:-4<x<4\right\}
\end{align}
  • b) \sqrt{3x+4} 2+\frac{3}{4}x-\frac{9}{64}{x}^{2}+\ldots\;\mbox{Valid for }\left\{x:-\frac{4}{3}<x<\frac{4}{3}\right\}
    • 
\begin{align}
\sqrt{3x+4}
&={\left( 4+3x\right)}^{\frac{1}{2}} \\
&=2{\left(1+\frac{3}{4}x\right)}^{\frac{1}{2}}\\
&=2\left[1+\left(\frac{1}{2}\right)\left(\frac{3}{4}x\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{\left(\frac{3}{4}x\right)}^{2} +\ldots \right]\\
&=2\left(1+\frac{3}{8}x-\frac{9}{128}{x}^{2}+\ldots \right) \\
&=2+\frac{3}{4}x-\frac{9}{64}{x}^{2}+\ldots  \\
\end{align}
    • \begin{align}
& \mbox{Valid for }\left| \frac{3}{4}x\right|<1, \therefore \left|x \right|<\frac{4}{3}\\
& \therefore \mbox{Valid for }\left\{x:-\frac{4}{3}<x<\frac{4}{3}\right\}
\end{align}
  • c) {\left(1+x\right)}^{2}\sqrt{1-2x} 1+x-\frac{3}{2}{x}^{2}+\ldots\;\mbox{Valid for }\left\{x:-\frac{1}{2}<x<\frac{1}{2}\right\}
    • 
\begin{align}
{\left(1+x\right)}^{2}\sqrt{1-2x}
&={\left(1+x\right)}^{2}{\left( 1-2x\right)}^{\frac{1}{2}} \\
&=\left(1+2x+{x}^{2}\right)\left[1+\left(\frac{1}{2}\right)\left(-2x\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{\left(-2x\right)}^{2}+\ldots \right] \\
&=\left(1+2x+{x}^{2}\right)\left(1-x-\frac{1}{2}{x}^{2}+\ldots \right) \\ 
&=1-x-\frac{1}{2}{x}^{2}+2x-2{x}^{2}+{x}^{2}+\ldots \\
&=1+x-\frac{3}{2}{x}^{2}+\ldots
\end{align}
    • \begin{align}
& \mbox{Valid for }\left| -2x\right|<1, \therefore \left|x \right|<\frac{1}{2}\\
& \therefore \mbox{Valid for }\left\{x:-\frac{1}{2}<x<\frac{1}{2}\right\}
\end{align}
  • d) \frac{2+x}{3-x} \frac{2}{3}+\frac{5}{9}x+\frac{5}{27}{x}^{2}+\ldots\;\mbox{Valid for }\left\{x:-3<x<3\right\}
    • 
\begin{align}
\frac{2+x}{3-x}
&=\left(2+x\right){\left(3-x\right)}^{-1}\\
&=\left(2+x\right)\left[ \frac{1}{3}{\left(1-\frac{x}{3}\right)}^{-1}\right]\\
&=\frac{1}{3}\left(2+x\right)\left[1+\left(-1\right)\left(-\frac{x}{3}\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(-\frac{x}{3}\right)}^{2}+\ldots\right]\\
&=\frac{1}{3}\left(2+x\right)\left(1+\frac{x}{3}+\frac{{x}^{2}}{9}+\ldots\right)\\
&=\frac{1}{3}\left(2+\frac{2}{3}x+\frac{2}{9}{x}^{2}+x+\frac{{x}^{2}}{3}+\ldots\right)\\
&=\frac{1}{3}\left(2+\frac{5}{3}x+\frac{5}{9}{x}^{2}+\ldots\right)\\
&=\frac{2}{3}+\frac{5}{9}x+\frac{5}{27}{x}^{2}+\ldots\\
\end{align}
    • \begin{align}
& \mbox{Valid for }\left| -\frac{x}{3}\right|<1, \therefore \left|x \right|<3\\
& \therefore\mbox{Valid for }\left\{x:-3<x<3\right\}
\end{align}
  • e) \sqrt[3]{\frac{1-x}{1+2x}} 1-x+{x}^{2}+\ldots\;\mbox{Valid for }\left\{x:-\frac{1}{2}<x<\frac{1}{2}\right\}
    • 
\begin{align}
\sqrt[3]{\frac{1-x}{1+2x}}
&={\left(1-x\right)}^{\frac{1}{3}}{\left(1+2x\right)}^{-\frac{1}{3}}\\
&=\left[1+\left(\frac{1}{3}\right)\left(-x\right)+\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}{\left(-x\right)}^{2}+\ldots\right]
\left[1+\left(-\frac{1}{3}\right)\left(2x\right)+\frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2!}{\left(2x\right)}^{2}+\ldots\right]\\
&=\left(1-\frac{1}{3}x-\frac{1}{9}{x}^{2}+\ldots \right)\left(1-\frac{2}{3}x+\frac{8}{9}{x}^{2}+\ldots \right)\\
&=1-\frac{2}{3}x+\frac{8}{9}{x}^{2}-\frac{1}{3}x+\frac{2}{9}{x}^{2}-\frac{1}{9}{x}^{2}+\ldots \\
&=1-x+{x}^{2}+\ldots
\end{align}
    • \begin{align}
& \mbox{Valid for }\left|-x\right|<1 \mbox{ and }\left|2x\right|<1 \\
& \therefore \left|x\right|<1 \mbox{ and }\left|x\right|<\frac{1}{2} \\
& \therefore \left|x\right|<\frac{1}{2}\\ 
& \therefore\mbox{Valid for }\left\{x:-\frac{1}{2}<x<\frac{1}{2}\right\}
\end{align}
  • f) \sqrt{\frac{{\left(1-2x\right)}^{3}}{4+x}} \frac{1}{2}-\frac{25}{16}x+\frac{243}{256}{x}^{2}+\ldots\;\mbox{Valid for }\left\{x:-\frac{1}{2}<x<\frac{1}{2}\right\}
    • 
\begin{align}
\sqrt{\frac{{\left(1-2x\right)}^{3}}{4+x}} & = {\left(1-2x\right)}^{\frac{3}{2}}{\left(4+x\right)}^{-\frac{1}{2}} \\
{\left(1-2x\right)}^{\frac{3}{2}} & =  1+\left(\frac{3}{2}\right)\left(-2x\right)+\frac{\left(\frac{3}{2}\right)\left(\frac{1}{2}\right)}{2!}{\left(-2x\right)}^{2}+\ldots \\
& =  1-3x+\frac{3}{2}{x}^{2}+\ldots \\
{\left(4+x\right)}^{-\frac{1}{2}} & =  \frac{1}{2}{\left(1+\frac{x}{4}\right)}^{-\frac{1}{2}} \\
& =  \frac{1}{2}\left[ 1+\left(-\frac{1}{2}\right)\left(\frac{x}{4}\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}{\left(\frac{x}{4}\right)}^{2}+\ldots\right] \\
& =  \frac{1}{2}-\frac{1}{16}x+\frac{3}{256}{x}^{2}+\ldots \\
\therefore \sqrt{\frac{{\left(1-2x\right)}^{3}}{4+x}} & =  \left(1-3x+\frac{3}{2}{x}^{2}+\ldots \right) \left(\frac{1}{2}-\frac{1}{16}x+\frac{3}{256}{x}^{2}+\ldots\right) \\
& = \frac{1}{2}-\frac{1}{16}x+\frac{3}{256}{x}^{2}-\frac{3}{2}x+\frac{3}{16}{x}^{2}+\frac{3}{4}{x}^{2}+\ldots \\
& = \frac{1}{2}-\frac{25}{16}x+\frac{243}{256}{x}^{2}+\ldots \\
\end{align}
    • \begin{align}
& \mbox{Valid for }\left|-2x\right|<1 \mbox{ and }\left|\frac{x}{4}\right|<1 \\
& \therefore \left|x\right|<\frac{1}{2}\mbox{ and }\left|x\right|<4\\
& \therefore \left|x\right|<\frac{1}{2}\\
&\therefore\mbox{Valid for }\left\{x:-\frac{1}{2}<x<\frac{1}{2}\right\}
\end{align}
  • g) \frac{1}{\left(2-x\right)\left(5+x\right)} \frac{1}{10}+\frac{3}{100}x+\frac{19}{1000}{x}^{2}+\ldots\;\mbox{Valid for }\left\{x:-2<x<2\right\}
    • 
\begin{align}
\frac{1}{\left(2-x\right)\left(5+x\right)} 
& = {\left(2-x\right)}^{-1}{\left(5+x\right)}^{-1} \\
& = \left[\frac{1}{2}{\left(1-\frac{x}{2}\right)}^{-1}\right]\left[\frac{1}{5}{\left(1+\frac{x}{5}\right)}^{-1}\right] \\
& = \frac{1}{10}\left[1+\left(-1\right)\left(-\frac{x}{2}\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(-\frac{x}{2}\right)}^{2}+\ldots\right] \left[1+\left(-1\right)\left(\frac{x}{5}\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(\frac{x}{5}\right)}^{2}+\ldots\right] \\
& = \frac{1}{10}\left(1+\frac{1}{2}x+\frac{1}{4}{x}^{2}+\ldots\right)\left(1-\frac{1}{5}x+\frac{1}{25}{x}^{2}+\ldots\right) \\
& = \frac{1}{10}\left(1-\frac{1}{5}x+\frac{1}{25}{x}^{2}+\frac{1}{2}x-\frac{1}{10}{x}^{2}+\frac{1}{4}{x}^{2}+\ldots\right) \\
& = \frac{1}{10}\left(1+\frac{3}{10}x+\frac{19}{100}{x}^{2}+\ldots \right) \\
& = \frac{1}{10}+\frac{3}{100}x+\frac{19}{1000}{x}^{2}+\ldots \\
\end{align}
    • \begin{align}
& \mbox{Valid for }\left|-\frac{x}{2}\right|<1 \mbox{ and }\left|\frac{x}{5}\right|<1\\
& \therefore \left|x\right|<2 \mbox{ and } \left|x\right|<5 \\
& \therefore \left|x\right|<2 \\
& \therefore \mbox{Valid for }\left\{x:-2<x<2\right\}
\end{align}

Ascending powers of 1/x

Note that {\left(1+x\right)}^{n} =1+nx+\frac{n\left(n-1\right)}{2!}{x}^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{x}^{3}+\ldots is in ascending powers of x\,

Whereas something like {\left(1-\frac{y}{2}\right)}^{n} will be in ascending powers of y\,

So how would we get an expansion which is in ascending powers of \frac{1}{x}? (Question will state clearly if it is required)

  • {\left(1+\frac{k}{x}\right)}^{n} =1+n\left( \frac{k}{x}\right)+\frac{n\left(n-1\right)}{2!}{\left( \frac{k}{x}\right)}^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{\left( \frac{k}{x}\right)}^{3}+\ldots

Note:

  • Thus we would need to change into the above from before expanding
  • Ascending powers of \frac{1}{x} can sometimes stated as descending powers of x\,, since \ldots\left(\frac{1}{x}\right)\ldots{\left(\frac{1}{x}\right)}^{2}\ldots{\left(\frac{1}{x}\right)}^{3} means \ldots{x}^{-1}\ldots{x}^{-2}\ldots{x}^{-3}

For example, if we are required to expand in ascending powers of \frac{1}{x}

  • \frac{1}{x+1} ={\left(x+1 \right)}^{-1}
    • Normally (ascending powers of x\,), we will just rearrange it to {\left(1+x \right)}^{-1}, but that won't work here.
    • Instead, we need to somehow have \frac{1}{x}
    • In this case, we force it by factoring an x\, out. \left(x+1 \right)=x\left(1+?\right) \left(x+1\right)=x\left(1+\frac{1}{x}\right)
    • \therefore {\left(x+1 \right)}^{-1}={\left[x\left(1+\frac{1}{x}\right)\right]}^{-1}
    • =\frac{1}{x}{\left(1+\frac{1}{x}\right)}^{-1}

Valid for : \left|\frac{1}{x}\right|<1 \therefore \left|x\right|>1

  • \frac{1}{2+x} ={\left(2+x \right)}^{-1}
    • ={\left(x+2 \right)}^{-1}
    • ={\left[x\left(1+\frac{2}{x}\right)\right]}^{-1}
    • =\frac{1}{x}{\left(1+\frac{2}{x}\right)}^{-1}

Valid for : \left|\frac{2}{x}\right|<1 \therefore \left|x\right|>2

Example

Express \frac{x-8}{\left(x+2\right)\left(x-3 \right)} in partial fractions. Hence, find it's expansion in ascending powers of \frac{1}{x} till the term in \frac{1}{{x}^{3}}. State the range where the expansion is valid.

    • First, using partial fractions (we will not show the working here), we will get
      • \frac{x-8}{\left(x+2\right)\left(x-3 \right)}=\frac{2}{x+2}-\frac{1}{x-3} Note : It is IMPORTANT to check that the partial fraction is correct before continuing
    • Noting that we are required to expand in ascending powers of \frac{1}{x}, let's expand the two terms separately first
      • \frac{2}{x+2} =2{\left(x+2 \right)}^{-1}
      • =2{\left[x\left(1+\frac{2}{x} \right) \right]}^{-1}
      • =\frac{2}{x}{\left(1+\frac{2}{x} \right)}^{-1}Not that since there is a \frac{2}{x} outside the bracket, we need to expand to {\left(\frac{1}{x} \right)}^{2} only
      • =\frac{2}{x}\left[1+\left(-1\right)\left(\frac{2}{x}\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(\frac{2}{x}\right)}^{2}+\ldots\right]
      • =\frac{2}{x}\left(1-\frac{2}{x}+\frac{4}{{x}^{2}}+\ldots \right)
      • =\frac{2}{x}-\frac{4}{{x}^{2}}+\frac{8}{{x}^{3}}+\ldots
      • \frac{1}{x-3} ={\left(x-3 \right)}^{-1}
      • ={\left[x\left(1-\frac{3}{x} \right) \right]}^{-1}
      • =\frac{1}{x}{\left(1-\frac{3}{x} \right)}^{-1}
      • =\frac{1}{x}\left[1+\left(-1\right)\left(-\frac{3}{x}\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(-\frac{3}{x}\right)}^{2}+\ldots\right]
      • =\frac{1}{x}\left(1+\frac{3}{x}+\frac{9}{{x}^{2}}+\ldots \right)
      • =\frac{1}{x}+\frac{3}{{x}^{2}}+\frac{9}{{x}^{3}}+\ldots
    • \therefore \frac{x-8}{\left(x+2\right)\left(x-3 \right)} =\left(\frac{2}{x}-\frac{4}{{x}^{2}}+\frac{8}{{x}^{3}}+\ldots\right)-\left(\frac{1}{x}+\frac{3}{{x}^{2}}+\frac{9}{{x}^{3}}+\ldots\right)
      • =\frac{1}{x}-\frac{7}{{x}^{2}}-\frac{1}{{x}^{3}}+\ldots
    • \mbox{Valid for}\, \left|\frac{2}{x}\right|<1\mbox{ and }\left|-\frac{3}{x}\right|<1
      • \therefore \left|x\right|>2\mbox{ and }\left|x\right|>3
      • \therefore \left|x\right|>3
      • \therefore x<-3\mbox{ or }x>3 \,

Finding value of n

When we are given the expansion (or some other information) and asked to find the value of n\,,

  • Just expand like normal, coefficients will be in terms of n\,
  • Compare with the given information and solve
  • Check that your final value satisfies the information


Example 1: If the first three terms of the expansion of {\left(1+ax\right)}^{n}is 1+x-\frac{1}{2}{x}^{2}, find the values of a\, and n\,.

  • {\left(1+ax\right)}^{n}
    • =1+n\left(ax\right)+\frac{n\left(n-1\right)}{2!}{\left(ax\right)}^{2}+\ldots
    • =1+nax+\frac{n\left(n-1\right){a}^{2}}{2}{x}^{2}+\ldots
    • \mbox{Comparing with }1+x-\frac{1}{2}{x}^{2}
    • \therefore na=1\quad\frac{\qquad}{}\left(1\right)\qquad\frac{n\left(n-1\right){a}^{2}}{2}= -\frac{1}{2}\quad\frac{\qquad}{}\left(2\right) We just compare the coefficient, DO NOT write nax=x\,
    • \mbox{From }\left(1\right), a=\frac{1}{n}\quad\frac{\qquad}{}\left(3\right)
    • \left(3\right)\to\left(2\right), \therefore \frac{n\left(n-1\right)}{2}{\left(\frac{1}{n}\right)}^{2}=-\frac{1}{2}
    • \frac{n-1}{n}  = -1, \therefore n = \frac{1}{2} \therefore a = \frac{1}{\left(\frac{1}{2}\right)}=2
    • Check :{\left(1+2x \right)}^{\frac{1}{2}}
=1+\left(\frac{1}{2}\right)\left(2x \right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{\left(2x \right)}^{2}+\ldots =1+x-\frac{1}{2}{x}^{2}+\ldots

Example 2: In the expansion of {\left(2+x \right)}^{n}, given the coefficient of the term in {x}^{2}\, is two times the coefficient of the term in {x}^{3}\,, find value of n\, which is a positive integer.

  • Analyze : Since n\, is positive integer, we CAN use the {\left(a+b\right)}^{n} formula first, rather than having to make the first term to be 1\,
  • {\left(2+x\right)}^{n} ={2}^{n}+\binom{n}{1}{2}^{n-1}x+\binom{n}{2}{2}^{n-2}{x}^{2}+\binom{n}{3}{2}^{n-3}{x}^{3}+\ldots
    • Alternatively, we can also use the {T}_{r+1}\, formula
    • But \binom{n}{2} & \binom{n}{3} are difficult to manipulate, since we don't know the value of n\,, thus we change it to the alternative form, either by simplifying the factorials or simply refer back to the formula for non positive integer n\,
    • ={2}^{n}+n{2}^{n-1}x+\frac{n\left(n-1\right)}{2!}{2}^{n-2}{x}^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{2}^{n-3}{x}^{3}+\ldots
    • Let's use the information given, keep in mind we only equate the coefficients, not the term itself (don't write the x\,)
    • \therefore \frac{n\left(n-1\right)}{2!}{2}^{n-2} =2\frac{n\left(n-1\right)\left(n-2\right)}{3!}{2}^{n-3} Let's deal with all the numbers first.
    • n\left(n-1\right)\left(\frac{{2}^{n-2}}{{2}^{n-3}}\right)\left(\frac{3!}{2!2}\right)=n\left(n-1\right)\left(n-2\right) But \frac{{2}^{n-2}}{{2}^{n-3}} ={2}^{\left(n-2\right)-\left(n-3\right)}=2
    • 3n\left(n-1\right)=n\left(n-1\right)\left(n-2\right)
    • n\left(n-1\right)\left(n-2\right)-3n\left(n-1\right)=0
    • n\left(n-1\right)\left[\left(n-2\right)-3\right]=0
    • n\left(n-1\right)\left(n-5\right)=0
    • \therefore n=0, n=1, n=5, \mbox{but }n\geq 3 \mbox{ to have term in }{x}^{3}
    • \therefore n=5
    • Check : \binom{5}{2}{2}^{3}=80,\binom{5}{3}{2}^{2}=40
    • Alternatively,
      • \binom{n}{2}{2}^{n-2}=2\left[\binom{n}{3}{2}^{n-3}\right]
      • \frac{n!}{\left(n-2\right)!2!}\left(\frac{{2}^{n-2}}{{2}^{n-3}} \right)=\frac{2n!}{\left(n-3\right)!3!}
      • \left(\frac{3!}{2!}\right)\left(2\right)=\frac{\left(n-2\right)!}{\left(n-3\right)!}
      • 3=n-2\,
      • n=5\,

Exercise 6

Notes

  • These are pretty hard question, but there are similar to actual STPM questions. You can try to redo the examples again on your own first to get the hang of it before attempting.

1) Express \frac{3{x}^{2}-3x+2}{\left(x-3\right)\left(3x+1 \right)} in partial fractions. Hence, or otherwise, find the first three terms in the expansion \frac{3{x}^{2}-3x+2}{\left(x-3\right)\left(3x+1 \right)} 1+\frac{2}{x-3}-\frac{1}{3x+1}

  • a) in ascending powers of x\, -\frac{2}{3}+\frac{25}{9}x-\frac{245}{27}{x}^{2}+\ldots\;\mbox{Valid for }\left\{x:-\frac{1}{3}<x<\frac{1}{3}\right\}
  • b) in ascending powers of \frac{1}{x} 1+\frac{5}{3x}+\frac{55}{9{x}^{2}}+\ldots\;\mbox{Valid for }\left\{x:x<-3,x>3\right\}

For each expansion, find the range of values of x\, such that the expansion is valid.

    • \left(x-3\right)\left(3x+1 \right)=3{x}^{2}-8x-3
    • \begin{array}{rl}
&\quad \;1 \\
3{x}^{2}-8x-3 & \overline{)\quad 3{x}^{2}-3x+2} \\ 
&\underline{-\left( 3{x}^{2}-8x-3\right)} \\
&\qquad \qquad 5x+5
\end{array}
    • \begin{align}
& \therefore \frac{3{x}^{2}-3x+2}{\left(x-3\right)\left(3x+1\right)}=1+\frac{5x+5}{\left(x-3\right)\left(3x+1 \right)}\\
& \mbox{Let} \; \frac{5x+5}{\left(x-3\right)\left(3x+1 \right)}=\frac{A}{x-3}+\frac{B}{3x+1}\\
& \therefore 5x+5 =A\left(3x+1 \right)+B\left(x-3\right)\\
& \mbox{when} \; x=3,\; 20=10A,\; \therefore A=2\\
& \mbox{when} \; x=-\frac{1}{3},\; \frac{10}{3}=-\frac{10}{3}B,\; \therefore B=-1\\
& \therefore \frac{5x+5}{\left(x-3\right)\left(3x+1 \right)}=\frac{2}{x-3}-\frac{1}{3x+1}\\
& \therefore \frac{3{x}^{2}-3x+2}{\left(x-3\right)\left(3x+1 \right)}=1+\frac{2}{x-3}-\frac{1}{3x+1}\\
\end{align}
  • a)
    • 
\begin{align}
1+\frac{2}{x-3}-\frac{1}{3x+1}
&= 1+2{\left(x-3\right)}^{-1}-{\left(3x+1\right)}^{-1}\\ 
2{\left(x-3\right)}^{-1} &= -2{\left(3-x\right)}^{-1}\\ 
&= -\frac{2}{3}{\left(1-\frac{x}{3}\right)}^{-1}\\ 
&= -\frac{2}{3}\left[1+\left(-1\right)\left(-\frac{x}{3}\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(-\frac{x}{3}\right)}^{2}+\ldots \right]\\ 
&=-\frac{2}{3}\left(1+\frac{1}{3}x+\frac{1}{9}{x}^{2}+\ldots \right)\\
&=-\frac{2}{3}-\frac{2}{9}x-\frac{2}{27}{x}^{2}+\ldots \\
{\left(3x+1\right)}^{-1} &= {\left(1+3x\right)}^{-1} \\
&=1+\left(-1\right)\left(3x\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(3x\right)}^{2}+\ldots \\
& =1-3x+9{x}^{2}+\ldots \\
\therefore \frac{3{x}^{2}-3x+2}{\left(x-3\right)\left(3x+1\right)}&=1+\left(-\frac{2}{3}-\frac{2}{9}x-\frac{2}{27}{x}^{2}+\ldots\right)-\left(1-3x+9{x}^{2}+\ldots\right)\\
&=-\frac{2}{3}+\frac{25}{9}x-\frac{245}{27}{x}^{2}+\ldots \\
\end{align}
    • \begin{align}
&\mbox{Valid for }\left|-\frac{x}{3}\right|<1\mbox{ and }\left|3x\right|<1\\
& \therefore \left|x\right|<3\mbox{ and }\left|x\right|<\frac{1}{3}\\
& \therefore \left|x\right|<\frac{1}{3}\\
& \therefore \mbox{Valid for }\left\{x:-\frac{1}{3}<x<\frac{1}{3}\right\}
\end{align}
  • b)
    • \begin{align}
1+\frac{2}{x-3}-\frac{1}{3x+1}
&= 1+2{\left(x-3\right)}^{-1}-{\left(3x+1\right)}^{-1}\\ 
2{\left(x-3\right)}^{-1} &= \frac{2}{x}{\left(1-\frac{3}{x}\right)}^{-1}\\ 
&= \frac{2}{x}\left[1+\left(-1\right)\left(-\frac{3}{x}\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(-\frac{3}{x}\right)}^{2}\ldots\right]\\ 
&= \frac{2}{x}\left(1+\frac{3}{x}+\frac{9}{{x}^{2}}+\ldots \right)\\
&= \frac{2}{x}+\frac{6}{{x}^{2}}+\frac{18}{{x}^{3}}+\ldots\\
{\left(3x+1\right)}^{-1} &= \frac{1}{3x}{\left(1+\frac{1}{3x}\right)}^{-1}\\ 
&= \frac{1}{3x}\left[1+\left(-1\right)\left(\frac{1}{3x}\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(\frac{1}{3x}\right)}^{2}\ldots\right]\\ 
&= \frac{1}{3x}\left(1-\frac{1}{3x}+\frac{1}{9{x}^{2}}+\ldots \right)\\
&= \frac{1}{3x}-\frac{1}{9{x}^{2}}+\frac{1}{{27x}^{3}}+\ldots\\
\therefore \frac{3{x}^{2}-3x+2}{\left(x-3\right)\left(3x+1\right)}&=1+\left(\frac{2}{x}+\frac{6}{{x}^{2}}+\frac{18}{{x}^{3}}+\ldots\right)-\left(\frac{1}{3x}-\frac{1}{9{x}^{2}}+\frac{1}{{27x}^{3}}+\ldots\right)\\
&=1+\frac{5}{3x}+\frac{55}{9{x}^{2}}+\ldots \\
\end{align}
    • \begin{align}
&\mbox{Valid for }\left|-\frac{3}{x}\right|<1\mbox{ and }\left|\frac{1}{3x}\right|<1\\
& \therefore \left|x\right|>3\mbox{ and }\left|x\right|>\frac{1}{3}\\
& \therefore \left|x\right|>3 \\
& \therefore \mbox{Valid for }\left\{x:x<-3,x>3\right\}
\end{align}

2) If the first three terms of the expansion of {\left(1+ax\right)}^{n} is 1-\frac{1}{3}x+\frac{1}{6}{x}^{2} find the values of a\, and n\,. n=-\frac{1}{2}, a=\frac{2}{3}

    • \begin{align}
&{\left(1+ax\right)}^{n}\\
&=1+n\left(ax\right)+\frac{n\left(n-1\right)}{2!}{\left(ax\right)}^{2}+\ldots\\
&=1+nax+\frac{n\left(n-1\right){a}^{2}}{2}{x}^{2}+\ldots\\
&\mbox{Comparing with }1-\frac{1}{3}x+\frac{1}{6}{x}^{2}\\
& \therefore na=-\frac{1}{3}\quad\frac{\qquad}{}\left(1\right)\qquad\frac{n\left(n-1\right){a}^{2}}{2}= \frac{1}{6}\quad\frac{\qquad}{}\left(2\right)\\
&\mbox{From }\left(1\right), a=-\frac{1}{3n}\quad\frac{\qquad}{}\left(3\right)\\
& \left(3\right)\to\left(2\right), \therefore \frac{n\left(n-1\right)}{2}{\left(-\frac{1}{3n}\right)}^{2}=\frac{1}{6}\\
& \frac{n-1}{n}  = \frac{3}{2}, \therefore n = -\frac{1}{2} \therefore a = -\frac{1}{3\left(-\frac{1}{2}\right)}=\frac{2}{3}
\end{align}

3) Find the value of n\,which is a positive integer, if

  • a) In the expansion {\left(2+x \right)}^{n} given the coefficient of the term in x\, is four times the coefficient of the term in {x}^{3}\,, n\, which is a positive integer. n=4\,
    • \begin{align}
& {\left(2+x\right)}^{n}={2}^{n}+\binom{n}{1}{2}^{n-1}x+\binom{n}{2}{2}^{n-2}{x}^{2}+\binom{n}{3}{2}^{n-3}{x}^{3}+\ldots\\
& ={2}^{n}+n{2}^{n-1}x+\frac{n\left(n-1\right)}{2!}{2}^{n-2}{x}^{2}+\frac{n\left(n-1\right)\left(n-2\right)}{3!}{2}^{n-3}{x}^{3}+\ldots\\
& \therefore n{2}^{n-1}=4\left[\frac{n\left(n-1\right)\left(n-2\right)}{3!}{2}^{n-3}\right]\\
& n\left(\frac{{2}^{n-1}}{{2}^{n-3}}\right)\left(\frac{3!}{4}\right)=n\left(n-1\right)\left(n-2\right)\\
& 6n=n\left(n-1\right)\left(n-2\right)\\
& n\left[\left(n-1\right)\left(n-2\right)-6\right]=0\\
& n\left({n}^{2}-3n-4\right)=0\\
& n\left(n-4\right)\left(n+1\right)=0\\
& \therefore n=0, n=-1, n=4\\
& \therefore n=4\\
\end{align}
  • b) In the expansion {\left(1+\frac{1}{4}x\right)}^{n} given the coefficient of the term in {x}^{4}\, is five times the coefficient of the term in {x}^{5}\,. n=8\,
    • \begin{align}
& {\left(1+\frac{1}{4}x\right)}^{n}, {T}_{r+1}=\binom {n}{r}{\left(\frac{1}{4}x\right)}^{r}\\
& \mbox{term in} \quad {x}^{4}=\binom {n}{4}{\left(\frac{1}{4}x\right)}^{4} \mbox {term in} \quad {x}^{5}=\binom {n}{5}{\left(\frac{1}{4}x\right)}^{5}\\
& \binom {n}{4}{\left(\frac{1}{4}\right)}^{4} =5\binom {n}{5}{\left(\frac{1}{4}\right)}^{5}\\
& \frac{n\left(n-1\right)\left(n-2\right)\left(n-3\right)}{4!}=\frac{5n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)}{5!}\left(\frac{1}{4}\right)\\
& 4n\left(n-1\right)\left(n-2\right)\left(n-3\right)=n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-4\right)\\
& n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left[\left(n-4\right)-4\right]=0\\
& n\left(n-1\right)\left(n-2\right)\left(n-3\right)\left(n-8\right)=0\\
& \therefore n=0,n=1,n=2,n=3,n=8\\
& \mbox{Since}\quad n\geq 5\therefore n=8\\
\end{align}