Problem - 4312
Show that $$\left(\sum_{k=0}^{\infty}x^k\right)^2=\sum_{k=0}^{\infty}(k+1)x^k$$
Let $\{a_k\}$ and $\{b_k\}$ be two constant sequences where $a_0=a_1=\cdots = 1$ and $b_0=b_1=\cdots=1$, i.e. $$\sum_{k=0}^{\infty}a_kx^k=\sum_{k=0}^{\infty}x^k\quad\text{and}\quad\sum_{k=0}^{\infty}a_kx^k=\sum_{k=0}^{\infty}x^k$$
Then we have $$\left(\sum_{k=0}^{\infty}x^k\right)^2=\left(\sum_{k=0}^{\infty}a_kx^k\right)\left(\sum_{k=0}^{\infty}b_kx^k\right)=\sum_{n=0}^{\infty}\left(\sum_{k=0}^{n}a_kb_kx^n\right)=\sum_{n=0}^{\infty}(n+1)x^k$$