I was trying to solve the summation:

> 
> $${\sum\_{k=1}^n k\binom{n}{k} = n2^{n-1}}$$
> 
> 
> 

I started something like:

$${\sum\_{k=0}^n \binom{n}{k} = 2^n}$$
$${\Rightarrow \sum\_{k=1}^n \binom{n}{k} = 2^n - 2^0 \qquad(1)}$$
Similarly,
$${\Rightarrow \sum\_{k=2}^n \binom{n}{k} = 2^n - 2^1 \qquad(2)}$$
$${\vdots}$$
$${\Rightarrow \sum\_{k=n}^n \binom{n}{k} = 2^n - 2^{n-1} \qquad(n)}$$

---

On adding the above n equations, we get,
$${\sum\_{k=1}^n k\binom{n}{k} = n2^n - (2^0 + 2^1 + \cdots + 2^{n-1})}$$
As the expression inside brackets on RHS is a GP, the sum comes up as,
$${S\_n = 2^0\left(\frac{2^n - 1}{2-1}\right) = 2^n - 1}$$
$${\Rightarrow \sum\_{k=1}^n k\binom{n}{k} = n2^n - 2^n + 1 \neq n2^{n-1}}$$

I've checked my solution many times and have searched for similar solutions [here](https://math.stackexchange.com/questions/123655/evaluate-and-prove-by-induction-sum-kn-choose-k-sum-frac1kk1) and [here](https://math.stackexchange.com/questions/7757/how-to-prove-this-binomial-identity-sum-r-0n-r-n-choose-r-n2n-1), but there was no answer using this method. What I am doing wrong?