How to find closedform expression of this series?
Solution 1
We have $G(x) = \displaystyle\sum_{n=0}^\infty a_n x^n = \sum_{n=0}^\infty n^2x^n$.
To find this, start with the geometric series:
$$\displaystyle\sum_{n=0}^\infty x^n = \frac{1}{1x}$$
Differentiate both sides and multiply by $x$:
$$\displaystyle\sum_{n=0}^\infty nx^n = \frac{x}{(1x)^2}$$
And one more time:
$$\displaystyle\sum_{n=0}^\infty n^2x^n = \frac{x(1+x)}{(1x)^3}$$
So $\displaystyle G(x) = \frac{x(1+x)}{(1x)^3}$.
Solution 2
Note that $$ n^2=\binom{n+1}{2}+\binom{n}{2}\tag{1}. $$ Also $$ \frac{1}{(1x)^k}=\sum_{n=0}^\infty \binom{k+n1}{k1}x^n\tag{2}. $$ for $k\geq 1$ (by repeatedly differentiating the geometric series or by the extended binomial theorem). In particular $$ \frac{1}{(1x)^3}=\sum_{n=0}^\infty \binom{n+2}{2}x^n\tag{3}. $$ Mutiply by $x^n$ and sum on $n$ in (1) to get that $$ \sum_{n=0}^\infty n^2x^n =\sum_{n=0}^\infty\binom{n+1}{2}x^n+\sum_{n=0}^\infty\binom{n}{2}x^n =\frac{x}{(1x)^3}+\frac{x^2}{(1x)^3} =\frac{x(1+x)}{(1x)^3}.\tag{4} $$ by (3).
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T.Meyer
Updated on July 23, 2022Comments

T.Meyer over 1 year
Find closedform expression of generating function of this series:
$<0,1,4,k^2,...>$
How can i find the expression?

Oussama Boussif over 6 yearsPossible duplicate of How can i find closedform expression of generating function of this series?
