Calculate $\infty \sum n = 0 n^{3} {(\frac{x + 1}{2})}^{n}$ $sum_{n=0}^{infty}n^3((x+1)/2)^n$ ?

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Calculate $\infty \sum n = 0 n^{3} {(\frac{x + 1}{2})}^{n}$ $sum_{n=0}^{infty}n^3((x+1)/2)^n$ ?

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Answer 1 · 2016-06-13T23:35:40

$f (x) = \frac{2 (1 + x) (13 + 10 x + x^{2})}{{(x - 1)}^{4}}$ $f(x) = (2 (1 + x) (13 + 10 x + x^2))/(x-1)^4$ for $- 3 < x < 1$ $-3 < x < 1$

Explanation:

The target series is $S = \infty \sum n = 0 n^{3} {(\frac{x + 1}{2})}^{n}$ $S = sum_{n=0}^{infty}n^3((x+1)/2)^n$ with generic term
$t = n^{3} {(\frac{x + 1}{2})}^{n}$ $t=n^3((x+1)/2)^n$

Let us introduce $S_{b} = \infty \sum n = 0 {(\frac{x + 1}{2})}^{n}$ $S_b = sum_{n=0}^{infty}((x+1)/2)^n$ with generic term $t_{b} = {(\frac{x + 1}{2})}^{n}$ $t_b=((x+1)/2)^n$

$S_{b}$ $S_b$ is convergent for $- 3 < x < 1$ $-3 < x < 1$ . Comparing terms we have

$D_{3} = \frac{d^{3}}{{d x}^{3}} t_{b} = \frac{n (n - 1) (n - 2)}{2^{3}} {(\frac{x + 1}{2})}^{n - 3} =$ $D_3=d^3/(dx^3)t_b=(n(n-1)(n-2))/(2^3)((x+1)/2)^{n-3} =$
$\frac{n^{3}}{2^{3}} {(\frac{x + 1}{2})}^{n - 3} - \frac{3 n^{2}}{2^{3}} {(\frac{x + 1}{2})}^{n - 3} + \frac{2 n}{2^{3}} {(\frac{x + 1}{2})}^{n - 3}$ $n^3/(2^3)((x+1)/2)^{n-3}-(3n^2)/(2^3)((x+1)/2)^{n-3}+(2n)/(2^3)((x+1)/2)^{n-3}$

then

$n^{3} {(\frac{x + 1}{2})}^{n} = 2^{3} {(\frac{x + 1}{2})}^{3} \frac{d^{3}}{{d x}^{3}} t + 3 n^{2} {(\frac{x + 1}{2})}^{n} - 2 n {(\frac{x + 1}{2})}^{n}$ $n^3((x+1)/2)^n = 2^3((x+1)/2)^3d^3/(dx^3)t+3n^2((x+1)/2)^n-2n((x+1)/2)^n$

Also

$D_{2} = \frac{d^{2}}{{d x}^{2}} t_{b} = \frac{n (n - 1)}{n} {(\frac{x + 1}{2})}^{n - 2}$ $D_2=d^2/(dx^2)t_b = (n(n-1))/n((x+1)/2)^{n-2}$ and analogously

$2^{2} {(\frac{x + 1}{2})}^{2} \frac{d^{2}}{{d x}^{2}} t_{b} = n^{2} {(\frac{x + 1}{2})}^{n} - n {(\frac{x + 1}{2})}^{n}$ $2^2((x+1)/2)^2d^2/(dx^2)t_b = n^2((x+1)/2)^n-n((x+1)/2)^n$

$2 (\frac{x + 1}{2}) \frac{d}{d x} t_{b} = n {(\frac{x + 1}{2})}^{n}$ $2((x+1)/2)d/(dx)t_b = n((x+1)/2)^n$

$2 (\frac{x + 1}{2}) D_{1} = n {(\frac{x + 1}{2})}^{n}$ $2((x+1)/2)D_1= n((x+1)/2)^n$

Putting all together

$S_{0} = \infty \sum n = 0 {2^{3} {(\frac{x + 1}{2})}^{3} D_{3} + 3 \times 2^{2} {(\frac{x + 1}{2})}^{2} D_{2} + 2 (\frac{x + 1}{2}) D_{1}}$ $S_0 = sum_{n=0}^{infty}{2^3 ((x + 1)/2)^3 D_3 + 3 xx 2^2 ((x + 1)/2)^2 D_2 + 2 ((x + 1)/2) D_1}$

or

$S_{o} = \frac{2 (1 + x) (13 + 10 x + x^{2})}{{(x - 1)}^{4}}$ $S_o= (2 (1 + x) (13 + 10 x + x^2))/(x-1)^4$ for $- \infty < x < 1$ $-infty < x < 1$

Attached is a figure with the comparison between $S$ $S$ and

$S_{0} = \frac{2 (1 + x) (13 + 10 x + x^{2})}{{(x - 1)}^{4}}$ $S_0 =(2 (1 + x) (13 + 10 x + x^2))/(x-1)^4$

Note:

The final expression for $S_{0}$ $S_0$ is obtained considering that $D_{k}, {k = 1, 2, 3}$ $D_k, {k = 1,2,3}$ is applied on the equivalence $S_{b} \equiv \frac{1}{1 - (\frac{x + 1}{2})}$ $S_b equiv 1/(1-((x+1)/2)$ for $- 3 < x < 1$ $-3 < x < 1$

so

$D_{1} = \frac{2}{{(x - 1)}^{2}}$ $D_1 = 2/(x-1)^2$
$D_{2} = - \frac{4}{{(x - 1)}^{3}}$ $D_2 = -4/(x-1)^3$
$D_{3} = \frac{12}{{(x - 1)}^{4}}$ $D_3 = 12/(x-1)^4$

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