Find the sum of binomial coefficients 
Calculate the value of the sum 
  $$
  \sum_{i = 1}^{100} i\binom{100}{i} = 1\binom{100}{1} + 
                                       2\binom{100}{2} + 
                                       3\binom{100}{3} + 
                                       \dotsb + 
                                       100\binom{100}{100}
$$

What I have tried: 
$$\begin{align}
  S &= 0\binom{100}{0}+1\binom{100}{1}+ \dotsb +99\binom{100}{99}+100\binom{100}{100} \\ \\
    &=100\binom{100}{100}+99\binom{100}{99}+ \dotsb +1\binom{100}{1}+0\binom{100}{0}
\end{align}$$
and I'm stuck here, I don't know if it's true or not, any help will be appreciated.

$\bf{My\; Solution::}$ Using $$\displaystyle (1+x)^n = \binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+..........+\binom{n}{n}x^n$$
Now Diff. both side w.r to $x\;,$ We get 
$$\displaystyle n(1+x)^{n-1} = \binom{n}{1}x+\binom{n}{2}\cdot 2x+\binom{n}{3}\cdot 3x^2+.........+\binom{n}{n}\cdot nx^{n-1}$$
Now Put $x=1$ and $n=100\;,$ We get 
$$\displaystyle 100\cdot 2^{99} = \binom{100}{1}\cdot 1+\binom{100}{2}\cdot 2+\binom{100}{3}\cdot 3+..........+\binom{100}{100}\cdot 100$$
Can anyone have any idea about combinatorial prove, If Yes Then plz explain here,
Thanks 
