# Question d1c04

Apr 10, 2017

$2 n \left(n + 1\right) + 1$

#### Explanation:

For $\left\mid x \right\mid < 1$ we have

${\lim}_{n \to \infty} \frac{1 - {x}^{n + 1}}{1 - x} = {\lim}_{n \to \infty} {\sum}_{k = 0}^{n} {x}^{k} = \frac{1}{1 - x}$

and also

$\frac{{d}^{2}}{{\mathrm{dx}}^{2}} \left(\frac{1}{1 - x}\right) = \frac{2}{1 - x} ^ 3 = {\sum}_{k = 2}^{\infty} k \left(k - 1\right) {x}^{k - 2}$

so, for $\left\mid x \right\mid < 1$

${\left(1 + x\right)}^{2} / {\left(1 - x\right)}^{3} = \frac{1}{2} \left({x}^{2} + 2 x + 1\right) {\sum}_{k = 2}^{\infty} k \left(k - 1\right) {x}^{k - 2} =$

$= \frac{1}{2} \left({\sum}_{k = 2}^{\infty} k \left(k - 1\right) {x}^{k} + 2 {\sum}_{k = 2}^{\infty} k \left(k - 1\right) {x}^{k - 1} + {\sum}_{k = 2}^{\infty} k \left(k - 1\right) {x}^{k - 2}\right)$ and for $k \ge 2$

(1+x)^2/(1-x)^3=1/2sum_(k=2)^oo((k(k-1)+2k(k+1)+(k+1)(k+2))x^k#

or for $k \ge 2$

${\left(1 + x\right)}^{2} / {\left(1 - x\right)}^{3} = {\sum}_{k}^{\infty} \left(2 k \left(k + 1\right) + 1\right) {x}^{k}$ so the coefficient for ${x}^{n}$

is

$2 n \left(n + 1\right) + 1$