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

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.

• Guessing a degree 4 polynomial from 6 terms may be considered a bit bold. As if every sequence starting with two identical terms were constant ... Commented Oct 28, 2018 at 10:56
• What Hagen von Eitzen said. The next terms could be anything Commented Oct 28, 2018 at 11:01
• The question is meaningless unless you are given the general term. There are an infinity of functions so that the first 6 integer values satisfy what's given in the summation, each giving different sums. Commented Oct 28, 2018 at 11:11

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

• Thanks a lot @Robert Z Sir.Your explanation taught me something new Commented Oct 28, 2018 at 13:33

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

• Thank you for you explanation @lab bhattacharjee Sir. Commented Oct 28, 2018 at 13:59