Binomial Expansion Part2
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Binomial Coefficient Revisited
Let's look at the first few binomial coefficients
- And the next one will be
Expansion for Non Positive integer n
- Valid for
At a glance, it seems to be the same formula we have already learnt for positive integer
, 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
here includes
- It is an infinite series (never ending), and we MUST write
at the end. Normally we would be told which term we need to expand until.
- We MUST use
form, there is no
form for non positive integer
- Expansion is only valid for
- For non positive integers,
does not exist (your calculator will give an error), thus we MUST use the other representation of the coefficients.
Of Apples & Oranges & Bananas
Actually, there are only two variables here,
(the
is fixed). Lets see how we are going to use the formula.
- So we always start with
- Followed by "
", which is the
- The following terms,
- the "
"
- the "
",
- the "
Calculating Coefficients
Lets say we have
,
- the coefficient
will become
- a better way to write it is to first understand
as series of steps
Note : Again, the pattern of the signs is either
- all positive
- all negative (except first term)
- positive and negative alternately
Examples
Expand the following in ascending powers of
until the term in
Try these on your own :
Answer :
Answer :
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
Changing Into (1+x)^n
In the formula
- it MUST be a
- "
" can be anything
So what do we do if the question is NOT a
?
Suppose we have
, we can change the
to a
by
What if we have
?
But the above is without any power. How will
affect it?
Remember that
In other words, when we want to bring the
out, we must
Try these
Multiple Expansions
Keep these in mind
- its always easier to
than
- it is very difficult to divide polynomial
Example
Let's put all these to work:
Expand the following in ascending powers of
, neglecting the terms in
and above
Comparison
As always, let's take a breather to compare the different things we have learnt thus far.
- For positive integer
Validity
Let's have a look at the formula again
The expansion is an infinite series, thus as
, we must have
Note :
For
as
to be true, we must have
Thus, the expansion is only true when
We say the expansion is valid only for
or
.
Important
- The "
" here refer to WHATEVER we use "
"
- Final answers are normally written in the
form, but the modulus form is useful to simplify
Simplifying the Validity Range
- Simplify the modulus form FIRST before changing to the other from
- Since
is always positive, we CAN multiply it across
is always positive (
actually)
- Combination
and
and
and
Identifying the x
The "
" that we are going to put into
is the same whatever "
" we use in the expansion. In other words
- Valid for
Examples
- Valid for
- Valid for
- Valid for
- Valid for
- Valid for
- Valid for
- Valid for
- Valid for
- Valid for
Stating the Validity
What form we state it depends on the form of the question
- Question : State the range of
where/such that the expansion is valid
- Answer : Valid for
- Question :State the set of values of
where/such that the expansion is valid
- Answer : Valid for
- Question :Find set
where the expansion is valid when
- Answer : Valid for
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
until the terms in
and above.
- a)
- b)
2) Expand the following in ascending powers of
, neglecting the terms in
and above. State the set of values of
where the expansion is valid.
- a)
- b)
- c)
- d)
- e)
- f)
- g)
- h)
Ascending powers of 1/x
Note that
is in
Whereas something like
will be in
So how would we get an expansion which is in ascending powers of
? (Question will state clearly if it is required)
Note:
- Thus we would need to change into the above from before expanding
- Ascending powers of
can sometimes stated as descending powers of
, since
means
For example, if we are required to expand in ascending powers of
Valid for :
Valid for :
Example
Express
in partial fractions. Hence, find it's expansion in ascending powers of
till the term in
. State the range where the expansion is valid.

Finding value of n
When we are given the expansion (or some other information) and asked to find the value of
,
Example 1:
If the first three terms of the expansion of
is
, find the values of
and
.
Example 2:
In the expansion of
, given the coefficient of the term in
is two times the coefficient of the term in
, find value of
which is a positive integer.
- Analyze :
- Alternatively,
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
in partial fractions. Hence, or otherwise, find the first three terms in the expansion
- a) in ascending powers of
- b) in ascending powers of
For each expansion, find the range of values of
such that the expansion is valid.
2) If the first three terms of the expansion of
is
find the values of
and
.
3) Find the value of
which is a positive integer, if
- a) In the expansion
given the coefficient of the term in
is four times the coefficient of the term in
,
which is a positive integer.
- b) In the expansion
given the coefficient of the term in
is five times the coefficient of the term in
.
















(but we might be asked to expand until certain terms only)




, which, though correct, is messy to calculate, not to mention creates a high risk of careless mistake


and 


earlier, thus we actually do one extra step only
yet.







using calculator, rather than typing the whole thing (waste time and easy to press wrongly), we might as well take
(product of numerators) divide by
(product of denominators) and then divide by
, and then determine the sign.
Note:
instead of





Note: 
Note: 

![\sqrt[3]{1+x}](/images/math/4/b/c/4bc632c84e56c1362337f9dad44ebd07.png)
Note: 





Now, the 

![={\left[2\left(1+2x\right)\right]}^{n}](/images/math/2/8/3/28316a8e6d93bb4728704333d34354ab.png)


![={\left[2\left(1+\frac{x}{2}\right)\right]}^{-1}](/images/math/6/7/d/67d942502e7635efe9717f5fbf681d3a.png)



![={\left[4\left(1+\frac{x}{4}\right)\right]}^{\frac{1}{2}}](/images/math/d/3/b/d3bb05936091a39f6ab3b983ab534ece.png)


![={\left[-\left(1-2x\right)\right]}^{-3}](/images/math/5/1/d/51d8f27b7e9e99602953654ad0371d2c.png)
Note: 
![\sqrt[3]{3x+27}](/images/math/e/5/2/e524240a5a33820c8651276ac4059be8.png)

![={\left[27\left(1+\frac{1}{9}x\right)\right]}^{\frac{1}{3}}](/images/math/4/a/6/4a65103751b36fc3542699bfd3ae6b3f.png)











Analyze : 
![={\left[9\left(1+\frac{x}{9}\right)\right]}^{\frac{1}{2}}](/images/math/5/3/7/537fee16287ea7de080741a157119996.png)
Now we can expand. What about the ![=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]](/images/math/c/8/a/c8ab202f4af279723d68946857317adf.png)


Analyze : 




and 
and 





or
form
(which is of course wrong). We can compare this with geometric series, where sum to infinity only exists if
, so that as
, 



(modulus takes care of the negative sign already)


or 


or 

(modulus sign is not needed anymore)







We must see from this step, right before we use the formula (rather than the earliest form)

The first bracket DON'T need since we did not expand it
The first bracket DON'T need since it's expansion for positive integer ![=\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]](/images/math/1/a/6/1a657bdd230669123c4bdb0c5987fae1.png)

(without set notation, marks won't be given)
(just a fancy way to ask you for the set)


























![\therefore {\left(x+1 \right)}^{-1}={\left[x\left(1+\frac{1}{x}\right)\right]}^{-1}](/images/math/0/2/9/02991e0d6f213ecc4f30d94d1b090f19.png)




![={\left[x\left(1+\frac{2}{x}\right)\right]}^{-1}](/images/math/9/8/1/9810c33b70b616e8756d32dba2c80e3a.png)






![=2{\left[x\left(1+\frac{2}{x} \right) \right]}^{-1}](/images/math/1/3/f/13fc602aead9f835bbbc940d47bef7fd.png)
Not that since there is a
outside the bracket, we need to expand to
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]](/images/math/1/0/f/10f66ce8b765f2c4872b12bf5c581975.png)




![={\left[x\left(1-\frac{3}{x} \right) \right]}^{-1}](/images/math/f/8/8/f88ecb63f82d8b3ad4455d547f402480.png)

![=\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]](/images/math/4/0/6/406a97ff3c1e9e8869efa1bd522e6546.png)










We just compare the coefficient, DO NOT write 




formula

Let's deal with all the numbers first.



![n\left(n-1\right)\left[\left(n-2\right)-3\right]=0](/images/math/7/d/2/7d2a7fa14c8f97aef012b5e92a94d88a.png)














![\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}](/images/math/6/f/4/6f44affd0bd44793445b69b0cd524744.png)

![\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}](/images/math/f/c/d/fcde7a713977b30956e13eae66335f5e.png)

