$\newcommand{\binomial}[2]{\left( \begin{array}{c} #1 \\ #2 \end{array} \right)}$
I have the following sum : $\displaystyle \sum_{k=1}^{n} \binomial{n}{k} = n2^{n-1}$
The $\binomial{n}{k}$ is a binomial coefficient with n over k. I must do a proof by induction of this.
I know that I must first show that if I set n=1, then it must be true. But after that what do I do ?
On
In the easiest proofs by induction, you must assume the formula that must be proved for $n$ and show that it is also true for $n+1$. This is done by writing the LHS for $n+1$, making some algebra to isolate the LHS for $n$ and then substitute.
LHS for $n+1$: $$\sum_{k=1}^{n+1}k\binom{n+1}k$$ Note that this sum has one more term that the one for $n$. Note also that $$\binom{n+1}k=\binom n{k-1}+\binom nk$$ All this together allow us to write the sum in terms of $n$ instead of $n+1$.
Can you finish?
Assume that the identity is true for $n$ .We'll prove it for $n+1$ .
Use the Pascal identity to split the sum : $$\sum_{k=1}^{n+1} k \binom{n+1}{k}=\sum_{k=1}^{n+1}k \left (\binom{n}{k}+\binom{n}{k-1} \right )=\sum_{k=1}^{n} k \binom{n}{k}+\sum_{k=0}^{n} (k+1) \binom{n}{k}=n 2^{n-1}+\sum_{k=0}^{n} k \binom{n}{k}+\sum_{k=0}^{n} \binom{n}{k}=n 2^{n-1}+n 2^{n-1}+2^n=(2n+2)2^{n-1}=(n+1)2^n$$ so the identity holds for all $n$ (for $n=1$ is obvious)