<s>[INST] 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.
 [/INST] Integration Approach
Since
$$
\begin{align}
\frac{\mathrm{d}}{\mathrm{d}x}\sum_{k=0}^n\frac1{k+2}\binom{n}{k}x^{k+2}
&=\sum_{k=0}^n\binom{n}{k}x^{k+1}\\
&=x(1+x)^n
\end{align}
$$
we have
$$
\begin{align}
\sum_{k=0}^n\frac1{k+2}\binom{n}{k}
&=\int_0^1x(1+x)^n\,\mathrm{d}x\\
&=\int_1^2(x-1)x^n\,\mathrm{d}x\\
&=\frac{2^{n+2}-1}{n+2}-\frac{2^{n+1}-1}{n+1}
\end{align}
$$

Pre-calculus Approach
lab bhattacharjee has already given a hint for this approach, but I was working on adding it so I will include it.
Since
$$
\binom{n}{k}=\frac{(k+2)(k+1)}{(n+2)(n+1)}\binom{n+2}{k+2}
$$
and
$$
\binom{n+1}{k+1}=\frac{k+2}{n+2}\binom{n+2}{k+2}
$$
we have
$$
\begin{align}
\sum_{k=0}^n\frac1{k+2}\binom{n}{k}
&=\frac1{(n+1)(n+2)}\sum_{k=0}^n(k+1)\binom{n+2}{k+2}\\
&=\frac1{(n+1)(n+2)}\left[\sum_{k=0}^n(k+2)\binom{n+2}{k+2}-\sum_{k=0}^n\binom{n+2}{k+2}\right]\\
&=\frac1{(n+1)(n+2)}\left[(n+2)\sum_{k=0}^n\binom{n+1}{k+1}-\sum_{k=0}^n\binom{n+2}{k+2}\right]\\
&=\frac1{(n+1)(n+2)}\left[(n+2)\left(2^{n+1}-1\right)-\left(2^{n+2}-(n+2)-1\right)\right]\\
&=\frac{2^{n+1}}{n+1}-\frac{2^{n+2}-1}{(n+1)(n+2)}
\end{align}
$$

Noting that $\frac1{(n+1)(n+2)}=\frac1{n+1}-\frac1{n+2}$ and $2^{n+2}=2\cdot2^{n+1}$, we see that the two approaches give the same answer.
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