# How to find closed-form expression of this series?

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## Solution 1

We have $G(x) = \displaystyle\sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty n^2x^n$.

$$\displaystyle\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$

Differentiate both sides and multiply by $x$:

$$\displaystyle\sum_{n=0}^\infty nx^n = \frac{x}{(1-x)^2}$$

And one more time:

$$\displaystyle\sum_{n=0}^\infty n^2x^n = \frac{x(1+x)}{(1-x)^3}$$

So $\displaystyle G(x) = \frac{x(1+x)}{(1-x)^3}$.

## Solution 2

Note that $$n^2=\binom{n+1}{2}+\binom{n}{2}\tag{1}.$$ Also $$\frac{1}{(1-x)^k}=\sum_{n=0}^\infty \binom{k+n-1}{k-1}x^n\tag{2}.$$ for $k\geq 1$ (by repeatedly differentiating the geometric series or by the extended binomial theorem). In particular $$\frac{1}{(1-x)^3}=\sum_{n=0}^\infty \binom{n+2}{2}x^n\tag{3}.$$ Mutiply by $x^n$ and sum on $n$ in (1) to get that $$\sum_{n=0}^\infty n^2x^n =\sum_{n=0}^\infty\binom{n+1}{2}x^n+\sum_{n=0}^\infty\binom{n}{2}x^n =\frac{x}{(1-x)^3}+\frac{x^2}{(1-x)^3} =\frac{x(1+x)}{(1-x)^3}.\tag{4}$$ by (3).

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