Question:

$\displaystyle\sum_{r=0}^n(-1)^r\binom{n}{r}\dfrac{1}{r+3}$

$\dfrac{2}{(n+1)(n+2)(n+3)}$

$\dfrac{n!}{(n+3)!}\cdot2$

$\dfrac{2}{n+3}$

$\dfrac{1}{n+3}$

Step-by-Step Solution

Key Concept: Beta function: $\sum(-1)^r\binom{n}{r}\frac{1}{r+3}=\int_0^1(1-t)^n t^2 dt=B(n+1,3)=\frac{2}{(n+1)(n+2)(n+3)}$

$=\frac{2}{(n+1)(n+2)(n+3)}$.

Correct Answer: 2