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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.

bormtolearn
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5 Answers5

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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.

ml0105
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  • Why we multiplied by 1+x ? – bormtolearn Dec 22 '14 at 22:23
  • Sorry. Multiply by $(1+x)^{3}$. Observe that $\frac{(1+x)^{3}}{(1-x)^{3}} = (1+x)^{3} \cdot (1 + \sum_{i=1}^{\infty} \binom{i+2}{i} x^{i})$. We use the geometric series identity to expand out the denominator. – ml0105 Dec 22 '14 at 22:35
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$\dfrac{(1+x)}{(1-x)^3}^3 =\dfrac{6}{(1-x)} - \dfrac{12}{(1-x)^2} + \dfrac{8}{(1-x)^3} -1=6\sum_{n=0}^\infty x^n-12\sum_{n=0}^\infty (n+1)x^n+4\sum_{n=0}^\infty (n+1)(n+2)x^n=1+\sum_{n=1}^\infty (6-12n-12+4n^2+12n+8)x^n=1 + \displaystyle\sum_{n=1}^{\infty} (4n^2+2)x^n$

Liza
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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.

Jack D'Aurizio
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    (+1) Sometimes I get so wrapped up in trying to derive an equation, I forget that it may be far simpler to verify it. – robjohn Dec 23 '14 at 23:49
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Hint: Differentiate the series: $\dfrac{1}{1-x} = 1+x+x^2+\cdots$, and differentiate it again.

DeepSea
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  • i tried solving by the way you guided. I am getting 6(1+x+x^2...) -12(1+2x+3x^2.....) -4(2.1+3.2x+4.3x^2....) -1

    this is not getting me to the desired result. Pls correct me.

     – bormtolearn Dec 22 '14 at 22:19
  • @bormtolearn I think You get $6(1+x+x^2...) -12(1+2x+3x^2.....) +4(2\cdot 1+3\cdot2x+4\cdot3x^2....) -1 $ becouse $\frac{-8}{(x-1)^3}=\frac{8}{(1-x)^3}$ – Liza Dec 22 '14 at 22:29
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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 $n\ge0$.

Community
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robjohn
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