MathsQuestions

A Level
Binomial (Non-integer)

1. Expand $(1+x)^{-2}$ up to and including the term in $x^3$.

$1 - 2x + 3x^2 - 4x^3$

2. Expand $\dfrac{1}{1+3x}$ up to and including the term in $x^3$.

$1 - 3x + 9x^2 - 27x^3$

3. Expand the following up to and including the term in $x^3$, and state the values of $x$ for which the expansion is valid:

$(1+2x)^{-1}$
$\sqrt{1-x}$

$1 - 2x + 4x^2 - 8x^3$, $|x| < \frac{1}{2}$
$1 - \dfrac{1}{2}x - \dfrac{1}{8}x^2 - \dfrac{1}{16}x^3$, $|x| < 1$

4. Expand the following up to and including the term in $x^3$, and state the values of $x$ for which the expansion is valid:

$\sqrt[3]{1-3x}$
$\dfrac{1}{1-0.5x}$

$1 - x - x^2 - \dfrac{5}{3}x^3$, $|x| < \frac{1}{3}$
$1 + \dfrac{1}{2}x + \dfrac{1}{4}x^2 + \dfrac{1}{8}x^3$, $|x| < 2$

5. Expand the following up to and including the term in $x^3$, and state the values of $x$ for which the expansion is valid:

$(3+x)^{-2}$
$\sqrt{4-3x}$

$\dfrac{1}{9} - \dfrac{2}{27}x + \dfrac{1}{27}x^2 - \dfrac{4}{243}x^3$, $|x| < 3$
$2 - \dfrac{3}{4}x - \dfrac{9}{64}x^2 - \dfrac{27}{512}x^3$, $|x| < \dfrac{4}{3}$

6. Expand the following up to and including the term in $x^3$, and state the values of $x$ for which the expansion is valid:

$\sqrt[3]{8+x}$
$\dfrac{1}{4+x}$

$2 + \dfrac{1}{12}x -\dfrac{1}{36}x^2 + \dfrac{5}{20736}x^3$, $|x| < 8$
$\dfrac{1}{4} - \dfrac{1}{16}x + \dfrac{1}{64}x^2 - \dfrac{1}{256}x^3$, $|x| < 4$

7. 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.

$6 + 4x^3$

8. Find the expansion of $\dfrac{1+3x}{(1+2x)^3}$ up to and including the term in $x^3$.

$1 - 3x + 6x^2 - 8x^3$

9. Find the expansion of $\sqrt{\dfrac{2+3x}{8}}$ up to and including the term in $x^2$.

$\dfrac{1}{2} + \dfrac{3}{8}x - \dfrac{9}{64}x^2$

10. Find the coefficient of $x^2$ in the expansion of $$\dfrac{2+x}{\sqrt{4-2x}}$$

$\dfrac{7}{32}$

11. 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.

$2.949576$

12. 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.

$1.9746810$

13. Find the first three terms, in ascending powers of $x$, in the binomial expansion of $$\dfrac{2-x}{\sqrt{1+x}}$$

$(2-x)(1-\frac{1}{2}x+\frac{3}{8}x^2) = 2 - 2x + \frac{5}{4}x^2$

14. By expanding $(81-16x)^{0.25}$ up to the term in $x^2$, find an approximation for $\sqrt[4]{80}$ to 7 decimal places.

$2.9906979$

15. Given $$(2+ax)^b \approx \dfrac{1}{2}-\dfrac{3}{4}x$$ find $a$ and $b$.

$a=3$, $b=-1$

16. Use the first 3 terms of the binomial expansion of $\sqrt[3]{8+3x}$ to find an exact approximation for $\sqrt[3]{9}$.

Expansion: $2+\dfrac{1}{4}x-\dfrac{1}{32}x^2$
Use $x = \frac{1}{3}$ to get $\dfrac{599}{288}$

17. Given $$(1+ax)^b \approx 1-6x+24x^2$$ find $a$ and $b$.

$a = 2$, $b = -3$

18.

Find the first 4 terms of the expansion of $\sqrt{1-2x}$.
Hence, by using $x = 0.01$, find an approximation for $\sqrt{2}$ to 9 decimal places.

$1 - x - \frac{1}{2}x^2 - \frac{1}{2}x^3$
$1.414213571$

19. 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.

$n = -2$, $b = 3$, $|x| < \dfrac{1}{3}$

20. In the expansion of $(3x+k)^{-2}$, the coefficient of $x$ is $4$ times as large as the coefficient of $x^2$. Find $k$.

$-18$

21. Find the first two non-zero terms, in ascending powers of $x$, of the expansion of $$\dfrac{1+6x-12x^2}{(1+2x)^3}$$

$1+136x^3$

22. 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.

$3.1621875$

23. 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.

$3.316624792$

24. 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.

$(1+8x)^{\frac{1}{2}} = 1 + 4x - 8x^2 + 32x^3$
$x = 0.01$, $\sqrt{1.08} = \sqrt{\frac{36\times3}{100}} = \frac{3}{5}\sqrt{3}$
Use $x = 0.01$, multiply by $\frac{5}{3}$