<s>[INST] 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.
 [/INST] You can compute the sum directly, using
Note that $$r\cdot \binom nr=r\cdot \frac {n!}{r!\cdot(n-r)!}=n\cdot\frac {(n-1)!}{(r-1)!\cdot(n-r)}=n\cdot\binom {n-1}{r-1}$$
This gives you a factor $100$ you can extract from every term leaving the sum from $\binom {99}0$ to $\binom {99}{99}$. The sum of such a complete set of binomial coefficients is well known - consider $(1+1)^{99}$
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