Sum of $n$ terms of the series: $30+144+420+960+1890+3360+\cdots$

Question

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.

Answer 1

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: $$\sum_{n=k}^N\binom{n}{k}=\binom{N+1}{k+1}.$$

Answer 2

Hint:

If $U(n)=n(n+1)(n+2)(n+4)=n(n+1)(n+2)(n+3)+n(n+1)(n+2)$

If $V(n)=n(n+1)\cdots(n+k),$

Now, $\underbrace{r(r+1)\cdots(r+k)(r+k+1)}-\underbrace{(r+1)\cdots(r+k)(r+k+1)(r+k+2)}$

$=(r+1)\cdots(r+k)(r+k+1)[r-(r+k+2)]$

$$\implies(k+2)\cdot V(r+1)=T(r)-T(r-1)$$ which is Telescoping with $T(m)=m(m+1)\cdots(m+k)(m+k+1)$

Put $r+1=1,2,\cdots,n-1,n$