Q:

Sum of the series $1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$

Find the sum of $n$ terms of following series:

$$1\cdot3\cdot2^2+2\cdot4\cdot3^2+3\cdot5\cdot4^2+\cdots$$

I was trying to use $S_n=\sum T_n$, but while writing $T_n$ I get a term having $n^4$ and I don't know $\sum n^4$. Is there any other way find the sum?

A:

We have
$$ n(n+1)^2(n+2) = 24\binom{n+2}{4}+12\binom{n+2}{3}\tag{1}$$
hence
$$ \sum_{n=1}^{N}n(n+1)^2(n+2) = 24\binom{N+3}{5}+12\binom{N+3}{4} \tag{2}$$
is a consequence of the Hockey Stick identity, leading to:
$$ \sum_{n=1}^{N}n(n+1)^2(n+2) = \color{red}{\frac{1}{10} N (1+N) (2+N) (3+N) (3+2 N)}.\tag{3}$$

