Expand ${\left(1+x\right)}^5$ using the Binomial Theorem.

Consider the power series $1-x+x^2-x^3+x^4-...$

By considering the ratio of consecutive terms, explain why this series is equal to ${\left(1+x\right)}^{-1}$ and state the values of $x$ for which this equality is true.

Differentiate the equation obtained part (b) and hence, find the first four terms in a power series for ${\left(1+x\right)}^{-2}$.

Repeat this process to find the first four terms in a power series for ${\left(1+x\right)}^{-3}$.

Hence, by recognising the pattern, deduce the first four terms in a power series for ${\left(1+x\right)}^{-n}$, $n\in {\mathbb{Z}}^{+}$.

By substituting $x=0$, find the value of $a_0$.

By differentiating both sides of the expression and then substituting $x=0$, find the value of $a_1$.

Repeat this procedure to find $a_2$ and $a_3$.

Hence, write down the first four terms in what is called the Extended Binomial Theorem for ${\left(1+x\right)}^q\text{,}q\in \mathbb{Q}$.

Write down the power series for $\frac{1}{1+x^2}$.

Hence, using integration, find the power series for $\text{arctan}x$, giving the first four non-zero terms.

$1+5x+10x^2+10x^3+5x^4+x^5$      M1A1

[2 marks]

It is an infinite GP with $a=1\text{,}r=-x$      R1A1

$S_{\mathrm{\infty }}=\frac{1}{1-\left(-x\right)}=\frac{1}{1+x}={\left(1+x\right)}^{-1}$      M1A1AG

[4 marks]

${\left(1+x\right)}^{-1}=1-x+x^2-x^3+x^4-...$

$-1{\left(1+x\right)}^{-2}=-1+2x-3x^2+4x^3-...$       A1

${\left(1+x\right)}^{-2}=1-2x+3x^2-4x^3+...$       A1

[2 marks]

$-2{\left(1+x\right)}^{-3}=-2+6x-12x^2+20x^3...$      A1

${\left(1+x\right)}^{-3}=1-3x+6x^2-10x^3...$      A1

[2 marks]

${\left(1+x\right)}^{-n}=1-nx+\frac{n\left(n+1\right)}{2\text{!}}{x}^2-\frac{n\left(n+1\right)\left(n+2\right)}{3\text{!}}{x}^3...$     A1A1A1

[3 marks]

$1^q=a_0\Rightarrow a_0=1$      A1

[1 mark]

$q{\left(1+x\right)}^{q-1}=a_1+2a_{2}x+3a_3{x}^2+...$       A1

$a_1=q$       A1

[2 marks]

$q\left(q-1\right){\left(1+x\right)}^{q-2}=1\times 2a_2+2\times 3a_{3}x+...$       A1

$a_2=\frac{q\left(q-1\right)}{2\text{!}}$       A1

$q\left(q-1\right)\left(q-2\right){\left(1+x\right)}^{q-3}=1\times 2\times 3a_3+...$       A1

$a_3=\frac{q\left(q-1\right)\left(q-2\right)}{3\text{!}}$       A1

[4 marks]

${\left(1+x\right)}^q=1+qx+\frac{q\left(q-1\right)}{2\text{!}}{x}^2+\frac{q\left(q-1\right)\left(q-2\right)}{3\text{!}}{x}^3...$     A1

[1 mark]

$\frac{1}{1+x^2}=1-x^2+x^4-x^6+...$    M1A1

[2 marks]

$\text{arctan}x+c=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...$    M1A1

Putting $x=0\Rightarrow c=0$        R1

So $\text{arctan}x=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+...$        A1

[4 marks]