Closed form expression of Infinite Summation Squared I am trying to simplify the following expression
$$
\left(\sum_{n=0}^{\infty} t^{2n}\right)^2
$$
I tried finding out a pattern for the coefficients by writing just the first three terms and squaring, i.e 
$$
(1+t^2+t^4)^2=1+2t^2+3t^4+2t^6+t^8
$$
and I notice that the more and more terms I square, there seems to be a pattern,i.e for 3 terms, the coefficients are $ 1,2,3,2,1$ for 4 terms it is $1,2,3,4,3,2,1$ and so on. But could someone please tell me how I can get a closed form summation expression with this?
 A: $${\frac {1}{1-t}}=\sum _{n=0}^{\infty }t^{n}\quad {\text{ for }}|t|<1$$
let $t\rightarrow t^2$
$${\frac {1}{1-t^2}}=\sum _{n=0}^{\infty }t^{2n}$$
$$({\frac {1}{1-t^2}})^2=(\sum _{n=0}^{\infty }t^{2n})^2$$
we have
$${\frac {1}{(1-t^2)^{2}}}=\sum _{n=1}^{\infty }nt^{2n-2}$$
so
$$(\sum _{n=0}^{\infty }t^{2n})^2=\sum _{n=1}^{\infty }nt^{2n-2}=\sum _{n=0}^{\infty }(n+1)t^{2n}$$
A: Well, if you really insist on writing out $(\sum t^{2n})^2$ as a new summation $\sum a_n t^n$, rather than simply using the other answers for a closed form expression, it is going to really on the binomial theorem / collecting terms in the appropriate manner. 
Looking at 
 $$\left(\sum_{n=0}^\infty t^{2n} \right)^2 = (1 + t^2 + t^4 + \cdots)(1+t^2 + t^4 + \cdots)$$
we want to observe how many ways there are to produce terms of the appropriate exponent:


*

*Constant terms: we can only pick $1$ from the first terms and $1$ from the second, i.e. $1\cdot 1$. 

*$t^{2k+1}$ terms: No possibilities--both expressions only include $t$ to even powers, so no multiplication will give odd exponents. 

*$t^2 terms$: we can pick $1\cdot t^2$ or $t^2 \cdot 1$, for a coefficient of 2

*$t^4 terms$: we can pick $1 \cdot t^4$, $t^2\cdot t^2$, or $t^4 \cdot 1$. Thus the coefficient is 3.

*$t^6 terms$: we can pick $1 \cdot t^6$, $t^2 \cdot t^4$, $t^4 \cdot t^2$, or $t^6 \cdot 1$. Coefficient is 4.

*$t^{2n} terms$: we can pick $1 \cdot t^{2n}$, $t^2 \cdot t^{2n-2}$, $\dots$, $t^{2n} \cdot 1$. There are $n+1$ of these. 
Thus the general formula would be 
$$\left(\sum_{n=0}^\infty t^{2n}\right)^2 = \sum_{n=0}^\infty (n+1)t^{2n}$$
Note that if we were presented with $(\sum a_n t^{2n})^2$ instead, the answer would be much messier and would involve a double sum: $(\sum a_n t^{2n} )^2= \sum_n \sum_j a_j t^{2n}$
A: \begin{align}
\left(\sum_{n=0}^\infty t^{2n}\right)^2&=\left(\sum_{j=0}^\infty t^{2j}\right)\left(\sum_{k=0}^\infty t^{2k}\right)
\\&=\sum_{j=0}^\infty \sum_{k=0}^\infty t^{2j+2k}
\\&=\sum_{n=0}^\infty \text{(# of ways of writing }2n\text{ as the sum of two non-negative even integers)}\cdot t^{2n}
\\&=\sum_{n=0}^\infty \text{(# of ways of writing }n\text{ as the sum of two non-negative integers)}\cdot t^{2n}
\\&=\sum_{n=0}^\infty (n+1)\, t^{2n}.
\end{align}
