50 followers problem

Lets count of different way. Supose that we have $50$ guys and we want to form with them a committee (this committee will have one president). There are $\binom{50}{k}$ ways to chose the committee where $1\leq k\leq50$ , and $k$ ways to select the president for each committe. $\therefore$ the sum we want is the total of committes. But is the same if we select first the president and then each guy can or cannot be in the committe, that is to say, $\boxed{50\times 2^{49}}$ .

$\therefore \dbinom{50}{1}+2\dbinom{50}{2}+3\dbinom{50}{3}+4\dbinom{50}{4}+\ldots+50\dbinom{50}{50}=50\times 2^{49}$

Evaluating $50\times 2^{49}= 28147497671065600$ and the sum of digits is $\boxed{73}$

This involves an identity, $\sum_{k=1}^{n} \binom{n}{k} = n2^{n-1}$ , which I proved in this wiki .