Q1
Answer
Expand $(1+x)^{-2}$ up to and including the term in $x^3$.
$1 - 2x + 3x^2 - 4x^3$
Q2
Answer
Expand $\dfrac{1}{1+3x}$ up to and including the term in $x^3$.
$1 - 3x + 9x^2 - 27x^3$
Q3
Answer
Expand the following up to and including the term in $x^3$, and state the values of $x$ for which the expansion is valid:
  1. $(1+2x)^{-1}$
  2. $\sqrt{1-x}$
  1. $1 - 2x + 4x^2 - 8x^3$, $|x| < \frac{1}{2}$
  2. $1 - \dfrac{1}{2}x - \dfrac{1}{8}x^2 - \dfrac{1}{16}x^3$, $|x| < 1$
Q4
Answer
Expand the following up to and including the term in $x^3$, and state the values of $x$ for which the expansion is valid:
  1. $\sqrt[3]{1-3x}$
  2. $\dfrac{1}{1-0.5x}$
  1. $1 - x - x^2 - \dfrac{5}{3}x^3$, $|x| < \frac{1}{3}$
  2. $1 + \dfrac{1}{2}x + \dfrac{1}{4}x^2 + \dfrac{1}{8}x^3$, $|x| < 2$
Q5
Answer
Expand the following up to and including the term in $x^3$, and state the values of $x$ for which the expansion is valid:
  1. $(3+x)^{-2}$
  2. $\sqrt{4-3x}$
  1. $\dfrac{1}{9} - \dfrac{2}{27}x + \dfrac{1}{27}x^2 - \dfrac{4}{243}x^3$, $|x| < 3$
  2. $2 - \dfrac{3}{4}x - \dfrac{9}{64}x^2 - \dfrac{27}{512}x^3$, $|x| < \dfrac{4}{3}$
Q6
Answer
Expand the following up to and including the term in $x^3$, and state the values of $x$ for which the expansion is valid:
  1. $\sqrt[3]{8+x}$
  2. $\dfrac{1}{4+x}$
  1. $2 + \dfrac{1}{12}x -\dfrac{1}{36}x^2 + \dfrac{5}{20736}x^3$, $|x| < 8$
  2. $\dfrac{1}{4} - \dfrac{1}{16}x + \dfrac{1}{64}x^2 - \dfrac{1}{256}x^3$, $|x| < 4$
Q7
Answer
Use the first 4 terms of the Binomial expansions to show that $$2\sqrt{1+4x}+\dfrac{4}{1+x} \approx a+bx^3$$ where $a$ and $b$ are constants to be found.
$6 + 4x^3$
Q8
Answer
Find the expansion of $\dfrac{1+3x}{(1+2x)^3}$ up to and including the term in $x^3$.
$1 - 3x + 6x^2 - 8x^3$
Q9
Answer
Find the expansion of $\sqrt{\dfrac{2+3x}{8}}$ up to and including the term in $x^2$.
$\dfrac{1}{2} + \dfrac{3}{8}x - \dfrac{9}{64}x^2$
Q10
Answer
Find the coefficient of $x^2$ in the expansion of $$\dfrac{2+x}{\sqrt{4-2x}}$$
$\dfrac{7}{32}$
Q11
Answer
By expanding $(9-6x)^{\frac{1}{2}}$ up to the term in $x^3$, find the value of $\sqrt{8.7}$ correct to 7 significant figures.
$2.949576$
Q12
Answer
By expanding $(8-3x)^{\frac{1}{3}}$ up to the term in $x^3$, find an estimate for $\sqrt[3]{7.7}$, giving your answer to 7 decimal places.
$1.9746810$
Q13
Answer
Find the first three terms, in ascending powers of $x$, in the binomial expansion of $$\dfrac{2-x}{\sqrt{1+x}}$$
$(2-x)(1-\frac{1}{2}x+\frac{3}{8}x^2) = 2 - 2x + \frac{5}{4}x^2$
Q14
Answer
By expanding $(81-16x)^{0.25}$ up to the term in $x^2$, find an approximation for $\sqrt[4]{80}$ to 7 decimal places.
$2.9906979$
Q15
Answer
Given $$(2+ax)^b \approx \dfrac{1}{2}-\dfrac{3}{4}x$$ find $a$ and $b$.
$a=3$, $b=-1$
Q16
Answer
Use the first 3 terms of the binomial expansion of $\sqrt[3]{8+3x}$ to find an exact approximation for $\sqrt[3]{9}$.
Expansion: $2+\dfrac{1}{4}x-\dfrac{1}{32}x^2$ <br> Use $x = \frac{1}{3}$ to get $\dfrac{599}{288}$
Q17
Answer
Given $$(1+ax)^b \approx 1-6x+24x^2$$ find $a$ and $b$.
$a = 2$, $b = -3$
Q18
Answer
  1. Find the first 4 terms of the expansion of $\sqrt{1-2x}$.
  2. Hence, by using $x = 0.01$, find an approximation for $\sqrt{2}$ to 9 decimal places.
  1. $1 - x - \frac{1}{2}x^2 - \frac{1}{2}x^3$
  2. $1.414213571$
Q19
Answer
In the expansion of $(1+bx)^n$, the coefficient of $x$ is $-6$ and the coefficient of $x^2$ is $27$. Find the values of $b$ and $n$ and state the range of values for $x$ for which the expansion is valid.
$n = -2$, $b = 3$, $|x| < \dfrac{1}{3}$
Q20
Answer
In the expansion of $(3x+k)^{-2}$, the coefficient of $x$ is $4$ times as large as the coefficient of $x^2$. Find $k$.
$-18$
Q21
Answer
Find the first two non-zero terms, in ascending powers of $x$, of the expansion of $$\dfrac{1+6x-12x^2}{(1+2x)^3}$$
$1+136x^3$
Q22
Answer
Use the first four terms in the binomial expansion of $$\dfrac{15}{\sqrt{1-x}}$$ and $x = 0.1$ to find an approximation for $\sqrt{10}$ to 7 decimal places.
$3.1621875$
Q23
Answer
By expanding $(1-x)^{\frac{1}{2}}$ up to the term in $x^3$, and using $x = 0.01$, find the value of $\sqrt{11}$ to 9 decimal places.
$3.316624792$
Q24
Answer
By expanding $(1+8x)^{\frac{1}{2}}$ up to the term in $x^3$, and using $x = 0.01$, find $\sqrt{3}$ to 6 decimal places.
$(1+8x)^{\frac{1}{2}} = 1 + 4x - 8x^2 + 32x^3$ <br> $x = 0.01$, $\sqrt{1.08} = \sqrt{\frac{36\times3}{100}} = \frac{3}{5}\sqrt{3}$ <br> Use $x = 0.01$, multiply by $\frac{5}{3}$