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?
 A: Hint:
$$\dfrac{r+1}{r+2}\binom nr=\dfrac{(r+1)^2}{(r+1)(r+2)}\binom nr$$
$$\begin{align}\dfrac{r+1}{r+2}\binom nr=\dfrac{(r+1)^2}{(r+1)(r+2)}\binom nr=\cdots=\dfrac{(r+1)^2}{(n+1)(n+2)}\binom{n+2}{r+2}\end{align}$$
Now $(r+1)^2=r^2+2r+1=(r+2)(r+1)-2(r+1)+1$
$$\begin{align}(r+1)^2\binom{n+2}{r+2}=\cdots=(n+2)(n+1)\binom nr-2(n+1)\binom{n+1}{r+1}+\binom{n+2}{r+2}\end{align}$$
Now use $$(a+b)^m=\sum_{r=0}^m\binom mra^{m-r}b^r$$ for $(1)$
A: Like you said,
$$
\sum_{r=0}^n\frac{r+1}{r+2}{n\choose r}x^r=\sum_{r=0}^n{n\choose r}x^r-\sum_{r=0}^n\frac{1}{r+2}{n\choose r}x^r=(1+x)^n-\sum_{r=0}^n\frac{1}{r+2}{n\choose r}x^r
$$
For the second sum, we want to use a sequence of derivatives/integrals/multiplications to get from $\sum_{r=0}^n{n\choose r}x^r=(1+x)^n$ to $\sum_{r=0}^n\frac{1}{r+2}{n\choose r}x^r$. The way I did this is to multiply by $x$, so that I get a factor of $x^{r+1}$, then integrating to get a factor of $1/(r+2)$, then dividing by $x^2$ to return the power of x to $r$. The computations:
$$\begin{aligned}
\sum_{r=0}^n{n\choose r}x^r&=(1+x)^n\\
\sum_{r=0}^n{n\choose r}x^{r+1}&=x(1+x)^n\\
\sum_{r=0}^n{n\choose r}\frac{x^{r+2}}{r+2}&=\int_0^xx'(1+x')^n\text{ d}x'\\
&=\left[\frac{x'(1+x')^{n+1}}{n+1}\right]_0^x-\frac1{n+1}\int_0^x(1+x')^{n+1}\text{ d}x'\\
&=\frac{x(1+x)^{n+1}}{n+1}-\frac{(1+x)^{n+2}-1}{(n+1)(n+2)}\\
&=\frac{(nx+x-1)(1+x)^{n+1}+1}{(n+1)(n+2)}\\
\sum_{r=0}^n{n\choose r}\frac{x^{r}}{r+2}&=\frac{(nx+x-1)(1+x)^{n+1}+1}{(n+1)(n+2)x^2}
\end{aligned}$$
A: 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 $$\frac{1}{(n+1)(n+2)}\sum_2^{n+2} (r-1)^2 \binom{n+2}{r} x^{r-2} \\=\frac{1}{(n+1)(n+2)}\sum_2^{n+2} r(r-1)\binom{n+2}{r} x^{r-2} - \frac{1}{(n+1)(n+2)}\sum_2^{n+2}(r-1)\binom{n+2}{r}x^{r-2} \\ = \frac{1}{(n+1)(n+2)}\frac{d^2}{dx^2}(1+x)^{n+2} - \frac{1}{(n+1)(n+2)} \frac{d}{dx}\frac{(1+x)^{n+2}-1-(n+2)x}{x} \\ \vdots$$
