Complex binomial series $$\sum_{r=0}^n \left[\frac{r+1}{r+2} (^n C_r) (x^r)\right] $$
Can someone help me evaluate this summation? I could solve till a certain extent but was then stuck. Would really help! so i started by differentiating $x(x+1)^n$ and got the coefficient of $(r+1)$. then I integrated the function I got above to get the coefficient of $(r+2)$ in the denominator but it was going a bit lengthy and i even got an incorrect expression at the end. so just wanted to know if there's any fallacy in my method or there is another elegant method too.
The summation can also be written as:
$(1+x)^n$ - $$\sum_{r=0}^n \frac {^nC_r}{r+2} $$
is there a way to evaluate the second summation?

Using $$\frac{r+1}{r+2} \binom{n}{r} = \frac{1}{n+1} \cdot\frac{(r+1)^2}{r+2} \binom{n+1}{r+1} = \frac{1}{(n+1)(n+2)}\cdot (r+1)^2\binom{n+2}{r+2}$$ and re-indexing, your sum is $$\frac{1}{(n+1)(n+2)}\sum_2^{n+2} (r-1)^2 \binom{n+2}{r} x^{r-2} \\=\frac{1}{(n+1)(n+2)}\sum_2^{n+2} r(r-1)\binom{n+2}{r} x^{r-2} - \frac{1}{(n+1)(n+2)}\sum_2^{n+2}(r-1)\binom{n+2}{r}x^{r-2} \\ = \frac{1}{(n+1)(n+2)}\frac{d^2}{dx^2}(1+x)^{n+2} - \frac{1}{(n+1)(n+2)} \frac{d}{dx}\frac{(1+x)^{n+2}-1-(n+2)x}{x} \\ \vdots$$
