Formula for summation series

Question

I don't know if there exists any formula for this given expression. Can anyone help?

$$ \sum_{i=1}^N \prod_{k=1}^T (i+k) $$

Answer 1

Note that: $$\sum_{i=1}^N \prod_{k=1}^T (i+k)=\sum_{i=1}^N (i+1)(i+2)\cdots(i+T)=\\ \sum_{i=1}^N \frac{(i+T)!}{i!}=T!\cdot \sum_{i=1}^N {i+T\choose T}=\\ T!\cdot\left(\sum_{i=0}^N {i+T\choose T}-1\right)=T!\cdot\left({N+T+1\choose T+1}-1\right).$$

Answer 2

Note that $$\displaystyle \prod_{k=1}^T (i+k) = \frac{1}{T+1} \{(i+T+1)-i\} \prod_{k=1}^T(i+k)$$

$$ = \frac{1}{T+1}\left\{ \prod_{k=1}^{T+1} (i+k) - \prod_{k=1}^{T+1} (i-1+k)\right \}$$

Therefore $$\sum_{i=1}^N \prod_{k=1}^T (i+k) = \frac{1}{T+1} \sum_{i=1}^N \left\{ \prod_{k=1}^{T+1} (i+k) - \prod_{k=1}^{T+1} (i-1+k)\right \} $$

$$= \frac{1}{T+1} \left\{ \prod_{k=1}^{T+1} (N+k) - \prod_{k=1}^{T+1} k \right \}$$

$$ = \frac{1}{T+1} \left \{ \frac{(N+T+1)!}{N!} - (T+1)! \right \}$$ $$ = T! \left \{ {N+T+1 \choose N} - 1 \right \}$$