Binomial Expansion Past Year

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Contents

Preparation

  • Make sure you have have mastered all the materials in the previous parts.
  • If you are using the material here to learn this topic (as opposed to just a revision to supplement your school lessons), its' best you take a day (or more) off after learning the previous parts.
  • In fact, try take a week off, and then revise back all the materials before trying the following questions. This will be a good practice for how you are going to revise for the actual STPM.
  • Find a comfortable place & comfortable time.
  • DO NOT do this while you are half-awake, or directly after a long day in school. Else, you will frustrate yourself. Trust me.
  • Saving trees is a good thing, but DO NOT do this (in fact, any of the exercises) on rough paper/recycled paper.
  • "ROUGH PAPER = ROUGH WORK = CARELESS MISTAKES = LOSS OF MARKS"
  • Know that you will face questions that you have NEVER SEEN which will require you to adapt on the spot.
  • Keep a clock or watch handy and give yourself the suggested time to complete it. Check your answers too during that time limit.
  • After finishing, check the answers given. If you made mistakes or couldn't find the solution, you can refer to the answers/answers with guidance.

Note

The list here isn't complete, as I don't have (yet) a complete set of past year papers, especially the older ones. I also have not listed the marks provided as I don't have complete information (and thus, the estimated time is really just an estimate). Nevertheless, the questions here should still provide a very good practice.

Questions

Estimated time : 3 hours

1) Find the term that does not contain x\, in the expansion {\left[{x}^{2}+\frac{1}{{x}^{3}}\right]}^{10} .


2) Find and simplify the term {x}^{-3}\, for the expansion {\left(\frac{4x}{5}-\frac{5}{2x}\right)}^{9}


3) Find the fourth term in the binomial expansion {\left(x+2y\right)}^{12} in descending powers of x\,. Find also the coefficient of the term involving {x}^{2}{y}^{10}\, in this expansion.


4) Determine the coefficient of {x}^{10}\, in the expansion {\left(1+x\right)}^{5}{\left(3-2x\right)}^{7}.


5) If the coefficient of x\, in the expansion {\left(4+x\right)}^{n} is the same as the coefficient of {x}^{3}\, in the expansion {\left(4+x\right)}^{n+1} , where n\, is a positive integer, find the value of n\, .With this value of n\,, write the expansion of {\left(4+x\right)}^{n}.


6) If \frac{y}{x} is so small that the powers of three and above of \frac{y}{x} can be ignored, show that {\left(\frac{x-y}{x+y}\right)}^{n}=1-2n\left(\frac{y}{x}\right)+2{n}^{2}{\left(\frac{y}{x}\right)}^{2}.


7) Find the expansion \frac{{\left(1+x\right)}^{\frac{1}{2}}}{{\left(1-2x\right)}^{2}} in ascending powers of x\, until and including the term in {x}^{3}\, . State the range of values of x\, such that the expansion is valid.


8) If x\, is so small that {x}^{3}\, and higher powers of x\, can be ignored, express \frac{{\left(1+x\right)}^{2}}{{\left(1-3x\right)}^{\frac{1}{3}}} in the form a+bx+c{x}^{2}\,, stating the numerical values of a,b\, and c\,.


9) Express \frac{3x+5}{\left(x+2\right)\left(2x-1 \right)} as partial fractions. Hence or otherwise, find the first 3 terms in the expansion \frac{3x+5}{\left(x+2\right)\left(2x-1 \right)} in ascending powers of \frac{1}{x}. State the range of values of x\, such that the expansion is valid.


10) Find the expansion {\left(1-3x\right)}^{\frac{1}{3}} in ascending powers of x\, until and including the term in {x}^{5}\,. State the range value of for which the expansion is valid. By taking x=\frac{1}{8}, evaluate \sqrt[3]{5} correct to three decimal places.


11)a) Find the expansion \sqrt[3]{\frac{1-2x}{{\left(1+3x\right)}^{2}}} in ascending powers of x\, until the term in {x}^{3}\,. State the set of values of x\, for which the expansion is valid.


b) If the expansion of {\left(1+ax\right)}^{n} in ascending powers of x\, until the term in {x}^{2}\, is 1+\frac{1}{15}x-\frac{1}{225}{x}^{2} , find the values of a\, and n\,.

Using these values of a\, and n\,, and taking x=0.08 \, , find the value of \sqrt[3]{127} correct to four decimal places.


12) Express {\left(\frac{1+x}{1+2x}\right)}^{\frac{1}{2}} as a series of ascending powers of x\, up to the term in {x}^{3}\,. By taking x=\frac{1}{30}, find \sqrt{62} correct to four decimal places.


13) Expand {\left(1+8x\right)}^{\frac{1}{2}} in ascending powers of x\, up to the term in {x}^{3}\,. By taking x=\frac{1}{100} , find \sqrt{3} correct to five decimal places.


14) Expand {\left(1-x\right)}^{\frac{1}{2}} in ascending powers of x\, up to the term in {x}^{3}\,. Hence, find the value of \sqrt{7}correct to five decimal places.


15) Expand {\left(1+x\right)}^{\frac{2}{3}} and \frac{1+ax}{1+bx} , where \left|b\right|<1 , in ascending powers of x\, up to the term in {x}^{3}\,. For each expansion, state the set of values of x\, for which the expansion is valid. Hence, find the set S\, such that the two expansions are valid for x \in S.

If two expansion are alike until the term in {x}^{2}\, ,

a) Determine the values of a\, and b\,

b) Use x=\frac{1}{8} to obtain the approximation \sqrt[3]{81}\approx\frac{212}{49}

c) Find, correct to 5 decimal places, the difference between the terms in {x}^{3}\, for the two expansion when x=\frac{1}{8}.

Answers

1) Find the term that does not contain x\, in the expansion {\left[{x}^{2}+\frac{1}{{x}^{3}}\right]}^{10} .

  • 210\,
    • \begin{align} &{T}_{r+1}=\binom{10}{r}{\left({x}^{2}\right)}^{10-r}{\left(\frac{1}{{x}^{3}}\right)}^{r}\\ & \mbox{Degree }x = 2\left(10-r\right)-3r=20-5r \\ & \mbox{For term that does not contain }x, 20-5r=0, \therefore r=4 \\ & \therefore \mbox{Term that does not contain }x=\binom{10}{4}{\left({x}^{2}\right)}^{6}{\left(\frac{1}{{x}^{3}}\right)}^{4}=210 \end{align}

2) Find and simplify the term {x}^{-3}\, for the expansion {\left(\frac{4x}{5}-\frac{5}{2x}\right)}^{9}

  • 10500{x}^{-3}\,
    • \begin{align} &{T}_{r+1}=\binom{9}{r}{\left(\frac{4x}{5}\right)}^{9-r}{\left(-\frac{5}{2x}\right)}^{r}\\ &\mbox{Degree }x = \left(9-r\right)-r=9-2r\\ &\mbox{For term in }{x}^{-3}, 9-2r=-3, \therefore 2r=12, r=6\\ &\therefore \mbox{Term in }{x}^{-3}=\binom{9}{6}{\left(\frac{4x}{5}\right)}^{3}{\left(-\frac{5}{2x}\right)}^{6}=10500{x}^{-3} \end{align}

3) Find the fourth term in the binomial expansion {\left(x+2y\right)}^{12} in descending powers of x\,. Find also the coefficient of the term involving {x}^{2}{y}^{10}\, in this expansion.

  • 760{x}^{9}{y}^{3};67584\,
    • \begin{align} & {T}_{r+1}=\binom{12}{r}{x}^{12-r}{\left(2y\right)}^{r} \\ & {T}_{4}=\binom{12}{3}{x}^{9}{\left(2y\right)}^{3}=760{x}^{9}{y}^{3}\\ &\mbox{Form in }{x}^{2}{y}^{10}, 12-r=2, \therefore r=10 \\ & \therefore \mbox{Term in }{x}^{2}{y}^{10}=\binom{12}{10}{x}^{2}{\left(2y\right)}^{10}=67584{x}^{2}{y}^{10}\\ & \therefore \mbox{Coefficient of term in }{x}^{2}{y}^{10}=67584 \end{align}

4) Determine the coefficient of {x}^{10}\, in the expansion {\left(1+x\right)}^{5}{\left(3-2x\right)}^{7}.

  • -608\,
    • \begin{align} & {\left(1+x\right)}^{5}{\left(3-2x\right)}^{7}\\ & =\left[\ldots+\binom{5}{3}{x}^{3}+\binom{5}{4}{x}^{4}+{x}^{5}\right]\left[\ldots+\binom{7}{5}{3}^{2}{\left(-2x\right)}^{5}+\binom{7}{6}\left(3\right){\left(-2x\right)}^{6}+{\left(-2x\right)}^{7}\right]\\ &=\left(\ldots+10{x}^{3}+5{x}^{4}+{x}^{5} \right)\left(\ldots-6048{x}^{5}+1344{x}^{6}-128{x}^{7} \right)\\ &\therefore \mbox{Term in }{x}^{10}=\left[-10\left(128\right)+5\left(1344\right)-6048\right]{x}^{10}=-608{x}^{10}\\ &\mbox{Coefficient of term in }{x}^{10}=-608 \end{align}

5) If the coefficient of x\, in the expansion {\left(4+x\right)}^{n} is the same as the coefficient of {x}^{3}\, in the expansion {\left(4+x\right)}^{n+1} , where n\, is a positive integer, find the value of n\, .With this value of n\,, write the expansion of {\left(4+x\right)}^{n}.

  • n=5;1024+1280x+640{x}^{2}+160{x}^{3}+20{x}^{4}+{x}^{5}\,
    • \begin{align} & {\left(4+x\right)}^{n}={4}^{n}+\binom{n}{1}{4}^{n-1}x+\ldots\\ & {\left(4+x\right)}^{n+1}={4}^{n+1}+\binom{n+1}{1}{4}^{n-1}x+\binom{n+1}{2}{4}^{n-2}{x}^{2}+\binom{n+1}{3}{4}^{n-3}{x}^{3}+\ldots\\ & \therefore \binom{n}{1}{4}^{n-1}=\binom{n+1}{3}{4}^{n-3}\\ & n\left( \frac{{4}^{n-1}}{{4}^{n-3}}\right)=\frac{\left(n+1\right)!}{\left(n-2\right)!3!}\\ & 4n=\frac{\left(n+1\right)n\left(n-1\right)\left(n-2\right)!}{\left(n-2\right)!3!}\\ & 4n=\frac{\left(n+1\right)n\left(n-1\right)}{3!}\\ & 24n=\left(n+1\right)n\left(n-1\right)\\ & n\left[\left(n+1\right)\left(n-1\right)-24\right]=0\\ & n\left({n}^{2}-25\right)=0\\ & n\left(n-5\right)\left(n+5\right)=0\\ & n=0,n=-5,n=5\\ & n \geq 3 \therefore n=5 \\ & {\left(4+x\right)}^{n}={4}^{5}+\binom{5}{1}{4}^{4}x+\binom{5}{2}{4}^{3}{x}^{2}+\binom{5}{3}{4}^{2}{x}^{3}+\binom{5}{4}4{x}^{4}+{x}^{5}\\ & \qquad \qquad=1024+1280x+640{x}^{2}+160{x}^{3}+20{x}^{4}+{x}^{5}\\ \end{align}

6) If \frac{y}{x} is so small that the powers of three and above of \frac{y}{x} can be ignored, show that {\left(\frac{x-y}{x+y}\right)}^{n}=1-2n\left(\frac{y}{x}\right)+2{n}^{2}{\left(\frac{y}{x}\right)}^{2}.

    • \begin{align} {\left(\frac{x-y}{x+y}\right)}^{n} &={\left[\frac{x\left(1-\frac{y}{x}\right)}{x\left(1+\frac{y}{x}\right)} \right]}^{n}\\ &={\left(1-\frac{y}{x}\right)}^{n}{\left(1+\frac{y}{x}\right)}^{-n} \\ &=\left[1+n\left(-\frac{y}{x}\right)+\frac{n\left(n-1\right)}{2!}{\left(-\frac{y}{x}\right)}^{2}+\ldots\right]\left[1+\left(-n\right)\left(\frac{y}{x}\right)+\frac{\left(-n\right)\left(-n-1\right)}{2!}{\left(\frac{y}{x}\right)}^{2}+\ldots\right]\\ &=\left[1-n\left(\frac{y}{x}\right)+\frac{n\left(n-1\right)}{2}{\left(\frac{y}{x}\right)}^{2}+\ldots\right]\left[1-n\left(\frac{y}{x}\right)+\frac{n\left(n+1\right)}{2}{\left(\frac{y}{x}\right)}^{2}+\ldots\right]\\ &=1-n\left(\frac{y}{x}\right)+\frac{n\left(n+1\right)}{2}{\left(\frac{y}{x}\right)}^{2}-n\left(\frac{y}{x}\right)+{n}^{2}{\left(\frac{y}{x}\right)}^{2}+\frac{n\left(n-1\right)}{2}{\left(\frac{y}{x}\right)}^{2}+\ldots \\ &=1-2n\left(\frac{y}{x}\right)+\left( \frac{{n}^{2}+1+2{n}^{2}+{n}^{2}-1}{2}\right){\left(\frac{y}{x}\right)}^{2}+\ldots\\ &=1-2n\left(\frac{y}{x}\right)+2{n}^{2}{\left(\frac{y}{x}\right)}^{2}+\ldots\\ \end{align}
    • \mbox{Since powers of three and above of }\frac{y}{x}\mbox{ can be ignored}
    • \therefore {\left(\frac{x-y}{x+y}\right)}^{n}=1-2n\left(\frac{y}{x}\right)+2{n}^{2}{\left(\frac{y}{x}\right)}^{2}

7) Find the expansion \frac{{\left(1+x\right)}^{\frac{1}{2}}}{{\left(1-2x\right)}^{2}} in ascending powers of x\, until and including the term in {x}^{3}\, . State the range of values of x\, such that the expansion is valid.

  • 1+\frac{9}{2}x+\frac{111}{8}{x}^{2}+\frac{601}{16}{x}^{3}+\ldots;-\frac{1}{2}<x<\frac{1}{2}
    • \begin{align} \frac{{\left(1+x\right)}^{\frac{1}{2}}}{{\left(1-2x\right)}^{2}} &={\left(1+x\right)}^{\frac{1}{2}}{\left(1-2x\right)}^{-2}\\ &=\left[ 1+\frac{1}{2}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\right] \left[1+\left(-2\right)\left(-2x\right)+\frac{\left(-2\right)\left(-3\right)}{2!}{\left(-2x\right)}^{2}+\frac{\left(-2\right)\left(-3\right)\left(-4\right)}{3!}{\left(-2x\right)}^{3}+\ldots\right]\\ &=\left(1+\frac{1}{2}x -\frac{1}{8}{x}^{2}+\frac{1}{16}{x}^{3}+\ldots\right)\left(1+4x+12{x}^{2}+32{x}^{3}+\ldots \right)\\ &=1+4x+12{x}^{2}+32{x}^{3}+\frac{1}{2}x+2{x}^{2}+6{x}^{3}-\frac{1}{8}{x}^{2}-\frac{1}{2}{x}^{3}+\frac{1}{16}{x}^{3}+\ldots \\ &=1+\frac{9}{2}x+\frac{111}{8}{x}^{2}+\frac{601}{16}{x}^{3}+\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 }-\frac{1}{2}<x<\frac{1}{2} \end{align}

8) If x\, is so small that {x}^{3}\, and higher powers of x\, can be ignored, express \frac{{\left(1+x\right)}^{2}}{{\left(1-3x\right)}^{\frac{1}{3}}} in the form a+bx+c{x}^{2}\,, stating the numerical values of a,b\, and c\,.

  • 1+3x+5{x}^{3};a=1,b=3,c=5\,
    • \begin{align} \frac{{\left(1+x\right)}^{2}}{{\left(1-3x\right)}^{\frac{1}{3}}} & ={\left(1+x\right)}^{2}{\left(1-3x\right)}^{-\frac{1}{3}} \\ & = \left(1+2x+{x}^{2}\right)\left[1+\left(-\frac{1}{3}\right)\left(-3x\right)+\frac{\left(-\frac{1}{3}\right)\left(-\frac{4}{3}\right)}{2!}{\left(-3x \right)}^{2}+\ldots \right] \\ & = \left(1+2x+{x}^{2}\right)\left(1+x+2{x}^{2}+\ldots\right)\\ & = 1+x+2{x}^{2}+2x+2{x}^{2}+{x}^{2}+\ldots \\ & = 1+3x+5{x}^{2}+\ldots \end{align}
    • \mbox{Since }{x}^{3}\mbox{ and higher powers of }x \mbox{ can be ignored},\,
    • \frac{{\left(1+x\right)}^{2}}{{\left(1-3x\right)}^{\frac{1}{3}}}=1+3x+5{x}^{2} \therefore a=1,b=3,c=5\,


9) Express \frac{3x+5}{\left(x+2\right)\left(2x-1 \right)} as partial fractions. Hence or otherwise, find the first 3 terms in the expansion \frac{3x+5}{\left(x+2\right)\left(2x-1 \right)} in ascending powers of \frac{1}{x}. State the range of values of x\, such that the expansion is valid.

  • \frac{1}{5\left(x+2\right)}+\frac{13}{5\left(2x-1\right)}; \frac{3}{2}\left(\frac{1}{x}\right)+\frac{1}{4}{\left(\frac{1}{x}\right)}^{2}+\frac{9}{8}{\left(\frac{1}{x}\right)}^{3}+\ldots; x<-2 \quad \mbox{or} \quad x>2
    • \begin{align} & \mbox{Let} \; \frac{3x+5}{\left(x+2\right)\left(2x-1 \right)} =\frac{A}{x+2}+\frac{B}{2x-1}\\ & \therefore 3x+5 =A\left(2x-1\right)+B\left(x+2\right)\\ & \mbox{when} \; x=-2,\; -1=-5A,\; \therefore A=\frac{1}{5}\\ & \mbox{when} \; x=\frac{1}{2},\; \frac{13}{2}=\frac{5}{2}B,\; \therefore B=\frac{13}{5}\\ & \therefore \frac{3x+5}{\left(x+2\right)\left(2x-1 \right)}=\frac{1}{5\left(x+2\right)}+\frac{13}{5\left(2x-1\right)}\\ \end{align}
    • \begin{align} \frac{1}{5\left(x+2\right)}+\frac{13}{5\left(2x-1\right)} & = \frac{1}{5}{\left(x+2\right)}^{-1}+\frac{13}{5}{\left(2x-1\right)}^{-1} \\ \frac{1}{5}{\left(x+2\right)}^{-1} &= \frac{1}{5x}{\left(1+\frac{2}{x}\right)}^{-1}\\ &= \frac{1}{5x}\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{1}{5x}\left(1-\frac{2}{x}+\frac{4}{{x}^{2}}+\ldots \right)\\ \frac{13}{5}{\left(2x-1\right)}^{-1} &= \frac{13}{10x}{\left(1-\frac{1}{2x}\right)}^{-1}\\ &= \frac{13}{10x}\left[1+\left(-1\right)\left(-\frac{1}{2x}\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(-\frac{1}{2x}\right)}^{2}+\ldots\right]\\ &= \frac{13}{10x}\left(1+\frac{1}{2x}+\frac{1}{4{x}^{2}}+\ldots \right)\\ \therefore \frac{1}{5\left(x+2\right)}+\frac{13}{5\left(2x-1\right)} & = \frac{1}{5x}\left(1-\frac{2}{x}+\frac{4}{{x}^{2}}+\ldots \right)+\frac{13}{10x}\left(1+\frac{1}{2x}+\frac{1}{4{x}^{2}}+\ldots \right) \\ & = \frac{1}{5x}-\frac{2}{5{x}^{2}}+\frac{4}{5{x}^{3}}+\frac{13}{10x}+\frac{13}{20{x}^{2}}+\frac{13}{40{x}^{3}}+\ldots\\ & = \frac{3}{2}\left(\frac{1}{x}\right)+\frac{1}{4}{\left(\frac{1}{x}\right)}^{2}+\frac{9}{8}{\left(\frac{1}{x}\right)}^{3}+\ldots\end{align}
    • \begin{align} & \mbox{Valid for }\left|\frac{2}{x}\right|<1\mbox{ and }\left|-\frac{1}{2x}\right|<1\\ & \therefore \left|x\right|>2 \mbox{ and }\left|x\right|>\frac{1}{2}, \therefore \left|x\right|>2 \\ & \therefore \mbox{Valid for }x<-2 \mbox{ or } x>2 \end{align}

10) Find the expansion {\left(1-3x\right)}^{\frac{1}{3}} in ascending powers of x\, until and including the term in {x}^{5}\,. State the range value of for which the expansion is valid. By taking x=\frac{1}{8}, evaluate \sqrt[3]{5} correct to three decimal places.

  • 1-x-{x}^{2}-\frac{5}{3}{x}^{3}-\frac{10}{3}{x}^{4}-\frac{22}{3}{x}^{5}+\ldots ; -\frac{1}{3}<x<\frac{1}{3};1.710
    • \begin{align} {\left(1-3x\right)}^{\frac{1}{3}} & = 1+\left(\frac{1}{3}\right)\left(-3x\right)+\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}{\left(-3x\right)}^{2} +\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}{\left(-3x\right)}^{3}+\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)\left(-\frac{8}{3}\right)}{4!}{\left(-3x\right)}^{4} +\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)\left(-\frac{8}{3}\right)\left(-\frac{11}{3}\right)}{5!}{\left(-3x\right)}^{5}+\ldots \\ & = 1-x-{x}^{2}-\frac{5}{3}{x}^{3}-\frac{10}{3}{x}^{4}-\frac{22}{3}{x}^{5}+\ldots \end{align}
    • \begin{align} &\mbox{Valid for }\left|-3x\right|<1, \therefore \left|x \right|<\frac{1}{3}\\ &\therefore \mbox{Valid for }-\frac{1}{3}<x<\frac{1}{3}\\ & \mbox{Substituting }x=\frac{1}{8}\\ &{\left[1-3\left( \frac{1}{8}\right)\right]}^{\frac{1}{3}}\approx 1-\frac{1}{8}-{\left( \frac{1}{8}\right)}^{2}-\frac{5}{3}{\left( \frac{1}{8}\right)}^{3}-\frac{10}{3}{\left( \frac{1}{8}\right)}^{4}-\frac{22}{3}{\left( \frac{1}{8}\right)}^{5}\\ &\mbox{L.H.S. }={\left[1-3\left( \frac{1}{8}\right)\right]}^{\frac{1}{3}} = \sqrt[3]{\frac{5}{8}}= \frac{1}{2}\sqrt[3]{5}\\ &\mbox{R.H.S. }=0.855082\\ &\therefore \frac{1}{2}\sqrt[3]{5} \approx 0.855082 \\ & \sqrt[3]{5}=1.710 \mbox{(Correct to 3 d.p.)}\\ \end{align}

11)a) Find the expansion \sqrt[3]{\frac{1-2x}{{\left(1+3x\right)}^{2}}} in ascending powers of x\, until the term in {x}^{3}\,. State the set of values of x\, for which the expansion is valid.

  • 1-\frac{8}{3}x+\frac{53}{9}{x}^{2}-\frac{1318}{81}{x}^{3}+\ldots; \mbox{Valid for} \; \left\{x: -\frac{1}{3}<x<\frac{1}{3}\right\}
    • \begin{align} \sqrt[3]{\frac{1-2x}{{\left(1+3x\right)}^{2}}} &={\left(1-2x\right)}^{\frac{1}{3}}{\left(1+3x\right)}^{-\frac{2}{3}}\\ &=\left[1+\left(\frac{1}{3}\right)\left(-2x\right)+\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)}{2!}{\left(-2x\right)}^{2}+\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}{\left(-2x\right)}^{3}+\ldots\right]\left [1+\left(-\frac{2}{3}\right)\left(3x\right)+\frac{\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{2!}{\left(3x\right)}^{2}+\frac{\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)\left(-\frac{8}{3}\right)}{3!}{\left(3x\right)}^{3}+\ldots\right]\\ &=\left(1-\frac{2}{3}x-\frac{4}{9}{x}^{2}-\frac{40}{81}{x}^{3}+\ldots \right)\left(1-2x+5{x}^{2}-\frac{40}{3}{x}^{3}+\ldots \right)\\ &=1-2x+5{x}^{2}-\frac{40}{3}{x}^{3}-\frac{2}{3}x+\frac{4}{3}{x}^{2}-\frac{10}{3}{x}^{3}-\frac{4}{9}{x}^{2}+\frac{8}{9}{x}^{3}-\frac{40}{81}{x}^{3} +\ldots \\ &=1-\frac{8}{3}x+\frac{53}{9}{x}^{2}-\frac{1318}{81}{x}^{3}+\ldots \end{align}
    • \begin{align} &\mbox{Valid for }\left|-2x\right|<1 \mbox{ and }\left|3x\right|<1 \\ & \therefore \left|x\right|<\frac{1}{2}\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) If the expansion of {\left(1+ax\right)}^{n} in ascending powers of x\, until the term in {x}^{2}\, is 1+\frac{1}{15}x-\frac{1}{225}{x}^{2} , find the values of a\, and n\,.

Using these values of a\, and n\,, and taking x=0.08 \, , find the value of \sqrt[3]{127} correct to four decimal places.

  • a=\frac{1}{5},n=\frac{1}{3};5.0265
    • \begin{align} &{\left(1+ax\right)}^{n}=1+n\left(ax\right)+\frac{n\left(n-1\right)}{2!}{\left(ax\right)}^{2}+\ldots\\ &\qquad \qquad =1+nax+\frac{n\left(n-1\right){a}^{2}}{2}{x}^{2}+\ldots \\ &\mbox{Comparing with }1+\frac{1}{15}x-\frac{1}{225}{x}^{2} \\ &\therefore na=\frac{1}{15}\quad\frac{\qquad}{}\left(1\right)\qquad\frac{n\left(n-1\right){a}^{2}}{2}= -\frac{1}{225}\quad\frac{\qquad}{}\left(2\right)\\ &\mbox{From }\left(1\right), a=\frac{1}{15n}\quad\frac{\qquad}{}\left(3\right)\\ &\left(3\right)\to\left(2\right), \therefore \frac{n\left(n-1\right)}{2}{\left(\frac{1}{15n}\right)}^{2}=-\frac{1}{225}\\ &\frac{n-1}{n} = -2, \therefore n = \frac{1}{3} \therefore a = \frac{1}{15\left(\frac{1}{3}\right)}=\frac{1}{5}\\ &\therefore {\left(1+\frac{1}{5}x\right)}^{\frac{1}{3}}=1+\frac{1}{15}x-\frac{1}{225}{x}^{2}+\ldots\\ &\mbox{Substituting }x=0.08 \\ &{\left[1+\frac{1}{5}\left(0.08\right)\right]}^{\frac{1}{3}} \approx 1+\frac{1}{15}\left(0.08\right)-\frac{1}{225}{\left(0.08\right)}^{2}\\ &\mbox{L.H.S. }={\left[1+\frac{1}{5}\left(0.08\right)\right]}^{\frac{1}{3}}= \sqrt[3]{\frac{127}{125}}= \frac{1}{5}\sqrt[3]{127}\\ &\mbox{R.H.S. }=1.0053049\\ &\therefore \frac{1}{5}\sqrt[3]{127} \approx 1.0053049\\ &\sqrt[3]{127}=5.0265\mbox{(Correct to 4 d.p.)} \end{align}

12) Express {\left(\frac{1+x}{1+2x}\right)}^{\frac{1}{2}} as a series of ascending powers of x\, up to the term in {x}^{3}\,. By taking x=\frac{1}{30}, find \sqrt{62} correct to four decimal places.

  • 1-\frac{1}{2}x+\frac{7}{8}{x}^{2}-\frac{25}{16}{x}^{3}+\ldots;7.8740
    • \begin{align} {\left(\frac{1+x}{1+2x}\right)}^{\frac{1}{2}} &={\left(1+x\right)}^{\frac{1}{2}}{\left(1+2x\right)}^{-\frac{1}{2}}\\ &=\left[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\right]\left [1+\left(-\frac{1}{2}\right)\left(2x\right)+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{2!}{\left(2x\right)}^{2}+\frac{\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)\left(-\frac{5}{2}\right)}{3!}{\left(2x\right)}^{3}+\ldots\right]\\ &=\left(1+\frac{1}{2}x-\frac{1}{8}{x}^{2}+\frac{1}{16}{x}^{3}+\ldots \right)\left(1-x+\frac{3}{2}{x}^{2}-\frac{5}{2}{x}^{3}+\ldots \right)\\ &=1-x+\frac{3}{2}{x}^{2}-\frac{5}{2}{x}^{3}+\frac{1}{2}x-\frac{1}{2}{x}^{2}+\frac{3}{4}{x}^{3}-\frac{1}{8}{x}^{2}+\frac{1}{8}{x}^{3}+\frac{1}{16}{x}^{3}+\ldots \\ &=1-\frac{1}{2}x+\frac{7}{8}{x}^{2}-\frac{25}{16}{x}^{3}+\ldots \end{align}
    • \begin{align} & \mbox{Substituting }x=\frac{1}{30}\\ & {\left[\frac{1+\left(\frac{1}{30}\right)}{1+2\left(\frac{1}{30}\right)}\right]}^{\frac{1}{2}}\approx1-\frac{1}{2}\left(\frac{1}{30}\right)+\frac{7}{8}{\left(\frac{1}{30}\right)}^{2}-\frac{25}{16}{\left(\frac{1}{30}\right)}^{3}\\ & \mbox{L.H.S. }={\left(\frac{1+\frac{1}{30}}{1+\frac{2}{30}}\right)}^{\frac{1}{2}}\\ & =\sqrt{\frac{\left({\frac{31}{30}}\right)}{\left(\frac{32}{30}\right)}}=\sqrt{\frac{31}{32}}=\sqrt{\frac{62}{64}}=\frac{1}{8}\sqrt{62}\\ &\mbox{R.H.S. }=0.984248\\ &\therefore \frac{1}{8}\sqrt{62} \approx 0.984248\\ &\sqrt{62}=7.8740\mbox{(Correct to 4 d.p.)} \end{align}

13) Expand {\left(1+8x\right)}^{\frac{1}{2}} in ascending powers of x\, up to the term in {x}^{3}\,. By taking x=\frac{1}{100} , find \sqrt{3} correct to five decimal places.

  • 1+4x-8{x}^{2}+32{x}^{3}+\ldots;1.73205\,
    • \begin{align} {\left(1+8x\right)}^{\frac{1}{2}} & = 1+\left(\frac{1}{2}\right)\left(8x\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{\left(8x\right)}^{2} +\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}{\left(8x\right)}^{3}+\ldots \\ & = 1+4x-8{x}^{2}+32{x}^{3}+\ldots \end{align}
    • \begin{align} & \mbox{Substituting }x=\frac{1}{100}\\ & {\left[1+8\left(\frac{1}{100}\right)\right]}^{\frac{1}{2}}\approx 1+4\left(\frac{1}{100}\right)-8{\left(\frac{1}{100}\right)}^{2}+32{\left(\frac{1}{100}\right)}^{3}\\ & \mbox{L.H.S. }={\left[1+8\left(\frac{1}{100}\right)\right]}^{\frac{1}{2}}=\sqrt{\frac{108}{100}}=\sqrt{\frac{27}{25}}=\sqrt{\frac{9\times3}{25}}=\frac{3}{5}\sqrt{3}\\ & \mbox{R.H.S. }=1.039232\\ & \therefore \frac{3}{5}\sqrt{3} \approx 1.039232 \\ & \sqrt{3}=1.73205\mbox{(Correct to 5 d.p.)} \end{align}

14) Expand {\left(1-x\right)}^{\frac{1}{2}} in ascending powers of x\, up to the term in {x}^{3}\,. Hence, find the value of \sqrt{7}correct to five decimal places.

  • 1-\frac{1}{2}x-\frac{1}{8}{x}^{2}-\frac{1}{16}{x}^{3}+\ldots;2.64575
    • \begin{align} {\left(1-x\right)}^{\frac{1}{2}} & = 1+\left(\frac{1}{2}\right)\left(-x\right)+\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)}{2!}{\left(-x\right)}^{2} +\frac{\left(\frac{1}{2}\right)\left(-\frac{1}{2}\right)\left(-\frac{3}{2}\right)}{3!}{\left(-x\right)}^{3}+\ldots \\ & = 1-\frac{1}{2}x-\frac{1}{8}{x}^{2}-\frac{1}{16}{x}^{3}+\ldots \end{align}
    • \begin{align} & \mbox{Substituting }x=\frac{1}{64}\\ & {\left(1-\frac{1}{64}\right)}^{\frac{1}{2}}\approx 1-\frac{1}{2}\left(\frac{1}{64}\right)-\frac{1}{8}{\left(\frac{1}{64}\right)}^{2}-\frac{1}{16}{\left(\frac{1}{64}\right)}^{3} \\ & \mbox{L.H.S. }={\left(1-\frac{1}{64}\right)}^{\frac{1}{2}}=\sqrt{\frac{63}{64}}=\sqrt{\frac{9 \times 7}{64}}=\frac{3}{8}\sqrt{7}\\ & \mbox{R.H.S. }=0.9921567\\ & \therefore \frac{3}{8}\sqrt{7} \approx 0.9921567 \\ & \sqrt{7}=2.64575\mbox{(Correct to 5 d.p.)} \end{align}

15) Expand {\left(1+x\right)}^{\frac{2}{3}} and \frac{1+ax}{1+bx} , where \left|b\right|<1 , in ascending powers of x\, up to the term in {x}^{3}\,. For each expansion, state the set of values of x\, for which the expansion is valid. Hence, find the set S\, such that the two expansions are valid for x \in S.

    • 1+\frac{2}{3}x-\frac{1}{9}{x}^{2}+\frac{4}{81}{x}^{3}+\ldots;\mbox{Valid for}\;\left\{x:-1<x<1\right\}
    • 1+\left(a-b\right)x+b\left(b-a\right){x}^{2}+{b}^{2}\left(a-b\right){x}^{3}+\ldots;\mbox{Valid for}\;\left\{x:-\frac{1}{\left|b\right|}<x<\frac{1}{\left|b\right|}\right\};
    • S=\left\{x:-1<x<1 \right\}

    • \begin{align} {\left(1+x\right)}^{\frac{2}{3}} & = 1+\left(\frac{2}{3}\right)x+\frac{\left(\frac{2}{3}\right)\left(-\frac{1}{3}\right)}{2!}{x}^{2} +\frac{\left(\frac{1}{3}\right)\left(-\frac{2}{3}\right)\left(-\frac{5}{3}\right)}{3!}{x}^{3}+\ldots \\ & = 1+\frac{2}{3}x-\frac{1}{9}{x}^{2}+\frac{4}{81}{x}^{3}+\ldots \end{align}
    • \begin{align} & \mbox{Valid for }\left|x\right|<1 \\ & \therefore \mbox{Valid for }\left\{ x:-1<x<1 \right\} \end{align}
    • \begin{align} \frac{1+ax}{1+bx} & = \left(1+ax\right){\left(1+bx\right)}^{-1}\\ & = \left(1+ax\right)\left[1+\left(-1\right)\left(bx\right)+\frac{\left(-1\right)\left(-2\right)}{2!}{\left(bx\right)}^{2}+\frac{\left(-1\right)\left(-2\right)\left(-3\right)}{3!}{\left(bx\right)}^{3}+\ldots \right] \\ & = \left(1+ax\right)\left(1-bx+{b}^{2}{x}^{2}-{b}^{3}{x}^{3}+\ldots \right) \\ & = 1-bx+{b}^{2}{x}^{2}-{b}^{3}{x}^{3}+ax-ab{x}^{2}+a{b}^{2}{x}^{3}+\ldots \\ & = 1+\left(a-b\right)x+b\left(b-a\right){x}^{2}+{b}^{2}\left(a-b\right){x}^{3}+\ldots \end{align}
    • \begin{align} & \mbox{Valid for }\left|bx\right|<1, \therefore \left|x \right|<\left|\frac{1}{b} \right|\\ & \therefore \mbox{Valid for }\left\{ x:-\frac{1}{\left|b\right|}<x<\frac{1}{\left|b\right|}\right\}\\ & \mbox{Since }\left|b\right|<1, \; \therefore \left|\frac{1}{b}\right|>1 \\ & \left|x\right|<1 \mbox{ and }\left|x \right|<\left|\frac{1}{b} \right|, \therefore \left|x\right|<1 \\ & \therefore S=\left\{x:-1<x<1 \right\} \end{align}

If two expansion are alike until the term in {x}^{2}\, ,

a) Determine the values of a\, and b\,

  • a=\frac{5}{6},b=\frac{1}{6}
    • \begin{align} &\mbox{Comparing }1+\frac{2}{3}x-\frac{1}{9}{x}^{2}\mbox{ with }1+\left(a-b\right)x+b\left(b-a\right){x}^{2}\\ & \therefore \frac{2}{3}=a-b\quad\frac{\qquad}{}\left(1\right)\qquad-\frac{1}{9}= b\left(b-a\right)\quad\frac{\qquad}{}\left(2\right)\\ &\frac{\left(2\right)}{\left(1\right)} : b = \frac{1}{6}, \therefore a = \frac{2}{3}+\frac{1}{6}=\frac{5}{6} \end{align}

b) Use x=\frac{1}{8} to obtain the approximation \sqrt[3]{81}\approx\frac{212}{49}

    • \begin{align} & \therefore {\left(1+x\right)}^{\frac{2}{3}} \approx \frac{1+\frac{5}{6}x}{1+\frac{1}{6}x} \\ &\mbox{Substituting }x=\frac{1}{8}\\ & {\left(1+\frac{1}{8}\right)}^{\frac{2}{3}} \approx \frac{1+\frac{5}{6}\left(\frac{1}{8}\right)}{1+\frac{1}{6}\left(\frac{1}{8}\right)}\\ & \mbox{L.H.S. }={\left(1+\frac{1}{8}\right)}^{\frac{2}{3}}=\sqrt[3]{{\left(\frac{9}{8}\right)}^{2}}=\sqrt[3]{\frac{81}{64}}=\frac{1}{4}\sqrt[3]{81}\\ & \mbox{R.H.S. }=\frac{53}{49}\\ & \therefore \frac{1}{4}\sqrt[3]{81} \approx \frac{53}{49}\\ & \therefore \sqrt[3]{81} \approx \frac{212}{49}\\ \end{align}

c) Find, correct to 5 decimal places, the difference between the terms in {x}^{3}\, for the two expansion when x=\frac{1}{8}.

  • 0.00006\,
  • \begin{align} \mbox{Difference between the terms in }{x}^{3} &= \frac{4}{81}{x}^{3}-{b}^{2}\left(a-b\right){x}^{3}\\ & =\frac{4}{81}{\left(\frac{1}{8}\right)}^{3}-{\left(\frac{1}{6}\right)}^{2}\left(\frac{5}{6}-\frac{1}{6}\right){\left(\frac{1}{8}\right)}^{3}\\ &=\frac{5}{82944}=0.00006\mbox{(Correct to 5 d.p.)} \end{align}

Answers(With guidance)

To Be Done

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