Verify that if $V_n = n(n+1)(n+2)\cdots(n+m)$ then $$V_{n+1} - V_n = (m+1)(n+1)(n+2)\cdots(n+m)$$ Given now that $U_n = (n+1)(n+2)\cdots(n+m)$ find sum of series $U_n$ from $N$ to $1$ in terms of $m$ and $N$.
I have done the first part of the questions. Having trouble with the sum part. Don't know how to approach it
Since $$V_{n+1}-V_n=\color{red}{(n+1)(n+2)\cdots (n+m)}(n+1+m)-n\color{red}{(n+1)(n+2)\cdots (n+m)}$$ $$=(n+1+m-n)\color{red}{(n+1)(n+2)\cdots (n+m)}=(m+1)\color{red}{U_n},$$ we have $$\begin{align}\sum_{k=1}^{N}U_k&=\frac{1}{m+1}\sum_{k=1}^{N}\left(V_{k+1}-V_{k}\right)\\&=\frac{1}{m+1}\left\{(V_2-V_1)+(V_3-V_2)+\cdots+(V_N-V_{N-1})+(V_{N+1}-V_N)\right\}\\&=\frac{1}{m+1}\left(V_{N+1}-V_1\right)\\&=\frac{1}{m+1}\left\{(N+1)(N+2)\cdots (N+1+m)-1\cdot 2\cdots (m+1)\right\}.\end{align}$$