I need to find the general term and the sum of $n$ terms of the series:$$30+144+420+960+1890+3360+\cdots$$
The answer provided my book is: $$U_n=n(n+1)(n+2)(n+4),\quad S_n=\frac{1}{20}n(n+1)(n+2)(n+3)(4n+21).$$
And I've no idea how to move on. It doesn't look like an arithmetic progression or a geometric progression. As far as I can tell it's not telescoping. What do I do? Any hints or solution will be appreciated.
Thanks in advance.
Sum of $n$ terms of the series: $30+144+420+960+1890+3360+\cdots$
sequences-and-seriessummationsummation-by-parts
Best Answer
Hint. Assuming that the general term is $U_n=n(n+1)(n+2)(n+4)$, we have that $$\begin{align} U_n&=n(n+1)(n+2)(n+3)+n(n+1)(n+2)=24\binom{n+3}{4}+6\binom{n+2}{3}. \end{align}$$ Then recall the Hockey-stick identity: $$\sum_{n=k}^N\binom{n}{k}=\binom{N+1}{k+1}.$$