How do I go about solving various summation of binomial coefficients like $\sum_{r=0}^{n} \binom{n}{r}f(r)$ I've come across many problems that require me to find summation of binomial coefficients. 
How go about solving these kind of summations of the form 
 $$\sum_{r=0}^{n} \binom{n}{r}f(r)$$
where f(r) is some function
for starters how to do these 


*

*$$\sum_{r=0}^{n} \binom{n}{r}(r+1)$$ 

*$$\sum_{r=0}^{n} \binom{n}{r}\frac{r^2}{3}$$
I remember someone using differentiation somewhere but I can't find where it was. So I thought this would be a good place where peiple can look 
 A: Use 
\begin{eqnarray*}
(1+x)^n = \sum_{r=0}^n \binom{n}{r} x^r 
\end{eqnarray*}
Differentiate and set $x=1$.
\begin{eqnarray*}
 \sum_{r=0}^n(r+1) \binom{n}{r} x^r \mid_{x=1} &=& \left( (1+x)^n +\frac{d}{dx} (1+x)^n \right)_{x=1} \\
 &=& \left( (1+x)^n +n (1+x)^{n-1} \right)_{x=1} \\
 &=& n2^{n-1}+2^n.
\end{eqnarray*}
\begin{eqnarray*}
 \sum_{r=0}^n r \binom{n}{r} x^r  &=&  n x(1+x)^{n-1}  \\
 \end{eqnarray*}
Differentiate again
\begin{eqnarray*}
 \sum_{r=0}^n r^2 \binom{n}{r} x^{r-1}  &=&  n(n-1) x(1+x)^{n-2}+n (1+x)^{n-1}  \\
 \end{eqnarray*}
Now set $x=1$ and we have 
\begin{eqnarray*}
 \sum_{r=0}^n \frac{r^2}{3} \binom{n}{r}   &=& \frac{1}{3} \left(n(n-1) 2^{n-2}+n 2^{n-1}\right)  \\
 \end{eqnarray*}
A: We know $(1+x)^n=\sum {n\choose r}x^r1^{n-r} $ differentiating with respect to $x $ we have $n (1+x)^{n-1}=\sum r {n\choose r}x^{r-1}(1) $putting $x=1$ we have  $n2^{n-1}=\sum r {n\choose r} $ if you differentiate first expression twice you can find answer to your second query.
A: Hint:
$$(r+1)\binom nr=r\binom nr+\binom nr$$
$$r^2\binom nr=r(r-1)\binom nr+r\binom nr$$
Now for $r>0,$ $$\displaystyle r\binom nr=\cdots=n\binom{n-1}{r-1}$$
Now for $r>1,$ 
$$r(r-1)\binom nr=\cdots=n(n-1)\binom{n-2}{r-2}$$
Now set $m=n,n-1,n-2$  in $$(1+1)^m=\sum_{r=0}^m\binom mr$$
