TITLE: $\sum_{i=0}^n (-1)^i\frac{C_i}{i+2}$ $C_i$ being the coefficient of $x^i$ in $(1+x)^n$
QUESTION [2 upvotes]: Evaluate: $\displaystyle\sum_{i=0}^n (-1)^i\frac{C_i}{i+2}$ where $C_i$ is the coefficient of $x^i$ in $(1+x)^n$
Attempt: We consider $f(x)=x(1-x)^n$ 
$\displaystyle\int_0^1x(1-x)^n  dx= \int_0^1x^{(2-1)}(1-x)^{(n+1)-1}  dx=B(2,n+1)=\frac{\Gamma(2)\Gamma(n+1)}{\Gamma(n+3)}= \frac{ n!}{(n+2)!}= \frac{1}{(n+1)(n+2)}=\frac{C_0}{2} - \frac{C_1}{3}+ \frac{C_2}{4}+...+ \frac{(-1)^nC_n}{n+2} $
Is this all?

REPLY [1 votes]: Hint:
Another way without calculus
$$\dfrac{\binom nr}{r+2}=\dfrac{(r+2-1)\binom nr}{(r+1)(r+2)}=\dfrac1{n+1}\binom{n+1}{r+1}-\dfrac1{(n+2)(n+1)}\binom{n+2}{r+2}$$
Now set $m=n+1,n+2$ in $$(1-1)^m=\sum_{r=0}^m(-1)^r\binom mr$$
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