Power series for the rational function $(1+x)^3/(1-x)^3$

Question

Show that $$\dfrac{(1+x)^3}{(1-x)^3} =1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$$

I tried with the partial frationaising the expression that gives me

$\dfrac{-6}{(x-1)} - \dfrac{12}{(x-1)^2} - \dfrac{8}{(x-1)^3} -1$ how to proceed further on this having doubt with square and third power term in denominator.

Answer 1

Note the identity:

$$\frac{1}{(1-x)^{r}} = \sum_{i=0}^{\infty} \binom{i + r - 1}{i} x^{i}$$

So you have $r = 3$, giving us:

$$\sum_{i=0}^{\infty} \binom{i + 2}{i} x^{i} = 1 + \sum_{i=1}^{\infty} \binom{i + 2}{i} x^{i}$$

Then we multiply by $(1 + x)^{3}$, the numerator:

$$(1 + x)^{3} \cdot (1 + \sum_{i=1}^{\infty} \binom{i + 2}{i} x^{i})$$

I leave the algebraic expansion to you.

Answer 2

Hint: Differentiate the series: $\dfrac{1}{1-x} = 1+x+x^2+\cdots$, and differentiate it again.

Answer 3

$\dfrac{(1+x)}{(1-x)^3}^3 =\dfrac{6}{(1-x)} - \dfrac{12}{(1-x)^2} + \dfrac{8}{(1-x)^3} -1=6\sum_{n=0}^\infty x^n-12\sum_{n=0}^\infty (n+1)x^n+4\sum_{n=0}^\infty (n+1)(n+2)x^n=1+\sum_{n=1}^\infty (6-12n-12+4n^2+12n+8)x^n=1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$

Answer 4

As shown in this answer, $\binom{-n\vphantom{1}}{k}=(-1)^k\binom{n+k-1}{k}$, therefore, $$ \begin{align} \frac{(1+x)^3}{(1-x)^3} &=(1+x)^3\sum_{k=0}^\infty\binom{-3}{k}(-x)^k\\ &=(1+x)^3\sum_{k=0}^\infty\binom{k+2}{k}x^k\\ &=\sum_{k=0}^\infty\sum_{j=0}^3\binom{3}{j}\binom{k-j+2}{k-j}x^k\\ &=1+\sum_{k=1}^\infty(4k^2+2)x^k \end{align} $$ since, for $k\ge1$, $$ \begin{align} &\overbrace{\binom{k+2}{k}}^{j=0}+\overbrace{3\binom{k+1}{k-1}}^{j=1}+\overbrace{3\binom{k}{k-2}}^{j=2}+\overbrace{\color{#C00000}{\binom{k-1}{k-3}}}^{j=3}\\[3pt] &=\binom{k+2}{2}+3\binom{k+1}{2}+3\binom{k}{2}+\color{#C00000}{\binom{k-1}{2}}\\[3pt] &=\frac{k^2+3k+2}2+\frac{3k^2+3k}2+\frac{3k^2-3k}2+\frac{k^2-3k+2}2\\[9pt] &=4k^2+2 \end{align} $$ Note that the red coefficients do not match for $k=0$; $\binom{-1}{-3}=0$ while $\binom{-1}{2}=1$. The rest match for $k=0$ and all match for $k\ge1$ because $\binom{n\vphantom{1}}{k}=\binom{n\vphantom{1}}{n-k}$ for $n\ge0$.

Answer 5

The most simple way to prove your identity, IMHO, is to multiply both sides by $(1-x)^3$.

This leads to: $$ 1+3x+3x^2+x^3\stackrel{?}{=}(1-3x+3x^2+x^3)\left(1+\sum_{n\geq 1}(4n^2+2)\,x^n\right).\tag{1}$$ If we set $a_n=(4n^2+2)$, for any $n\geq 4$ the coefficient of $x^n$ in the RHS is given by $a_n-3a_{n-1}+3a_{n-2}+a_{n-3}$ that is zero, since we are applying three times the backward difference operator to a polynomial in $n$ having degree two. So we just have to check that the first four coefficients, $[x^0],[x^1],[x^2],[x^3]$, match.