Summation of falling factorials

Question

I just want to know if I should evaluate $\sum(t+1)^\underline{4}$ the way we evaluate $\sum{t^\underline{4}}$.

Thanks.

Answer 1

Just shift $t$:

$$\sum_{t=1}^{n-1}(t+1)^{\underline 4}=\sum_{t=2}^nt^{\underline 4}=\frac15\left((n+1)^{\underline5}-2^{\underline 5}\right)=\frac15(n+1)^{\underline 5}$$

In effect I’m substituting $s=t+1$, rewriting the summation in terms of $s$, and then renaming $s$ back to $t$.