Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$ I am studying the book Equations and Inequalities by Herman et al, and am stuck on the following exercise:
Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$ 
The hint says to consider $(n+2)(n+1)S$.
Doing so, I get:
$(n+2)(n+1)S = \frac{1}{2}{n+2 \choose 0} + \frac{1}{3}{n+2 \choose 1} + \cdots + \frac{1}{n+2}{n+2 \choose n}$,
but I don't see how this is any easier to solve.
I wonder if somebody might help by expanding on the given hint, and maybe offering their suggestions as to how to use it properly.
Thanks.

$$\dfrac1{k+2}\binom nk=\dfrac{k+1}{(n+1)(n+2)}\cdot\dfrac{(n+2)!}{(k+2)!\{n+2-(k+2)\}!}=\dfrac{k+1}{(n+1)(n+2)}\cdot\binom{n+2}{k+2}$$
Now $\displaystyle(k+1)\cdot\binom{n+2}{k+2}$
$\displaystyle=(k+2-1)\cdot\binom{n+2}{k+2}$
$\displaystyle=(n+2)\cdot\binom{(n+1)!}{(k+1)!\{n+1-(k+1)\}!}-\binom{n+2}{k+2}$
$\displaystyle=(n+2)\cdot\binom{n+1}{k+1}-\binom{n+2}{k+2}$
Now use $\displaystyle(1+1)^m=\sum_{r=0}^m\binom mr$
