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

How do I prove this using mathematical induction?

Answer: 
To prove by induction, you call $P(n)$ the property $\sum\_{k = 1}^{n}k\binom{n}{k}=n2^{n-1}$. Then, you have to prove

* $P(1)$ is true
* if $P(n)$ is true for $n\geq1$, then $P(n+1)$ is also true

Then by induction $P(n)$ is true for all $n\geq1$.

Here (and often, but not always) $P(1)$ is easy, it amounts to $1=1$.

If $P(n)$ is true, then $\sum\_{k = 1}^{n}k\binom{n}{k}=n2^{n-1}$, hence

$$\sum\_{k=1}^{n+1}k\binom{n+1}{k}=\sum\_{k=1}^{n+1}k\binom{n}{k}+\sum\_{k=1}^{n+1}k\binom{n}{k-1}$$
$$=\sum\_{k=1}^{n+1}k\binom{n}{k}+\sum\_{k=0}^{n}(k+1)\binom{n}{k}$$
$$=\sum\_{k=1}^{n+1}k\binom{n}{k}+\sum\_{k=1}^{n}k\binom{n}{k}+\sum\_{k=0}^{n}\binom{n}{k}$$
$$=n2^{n-1}+(n+1){n \choose n+1}+n2^{n-1}+2^n$$
$$=n2^n+2^n=(n+1)2^n$$

That is, if $P(n)$ is true then

$$\sum\_{k=1}^{n+1}k\binom{n+1}{k}=(n+1)2^n$$

And $P(n+1)$ is true. Since $P(1)$ is also true, your formula is proved for all $n\geq1$.

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In the proof, I am using the well known sum

$$\sum\_{k=0}^n{n\choose k}=2^n$$

There is an elementary combinatorial proof mentioned in the comment below, and it's a perfectly valid mathematical proof (though in my student life I have encountered teachers who didn't like combinatorial proofs). Here is another proof, by induction.

Let $P(n)$ the property $\sum\_{k=0}^n{n\choose k}=2^n$. $P(0)$ is ibvious, as it amounts to $1=1$. Let's assume $P(n)$ is true for some $n\geq0$, then

$$\sum\_{k=0}^{n+1}{n+1\choose k}=1+\sum\_{k=1}^{n+1}{n+1\choose k}=1+\sum\_{k=1}^{n+1}\left[{n\choose k}+{n\choose k-1}\right]\\
=1+\sum\_{k=1}^{n}{n\choose k}+\sum\_{k=1}^{n+1}{n\choose k-1}\\
=\sum\_{k=0}^{n}{n\choose k}+\sum\_{k=0}^{n}{n\choose k}\\
=2^{n+1}$$

And $P(n+1)$ is true, hence by induction, $P(n)$ is true for all $n\geq0$.

Notice that in the proof I have used the fact that ${n\choose n+1}=0$.
I am sure you know of the binomial expansion$$(x+1)^n=\sum\_{k=0}^n\binom nkx^k$$ If you differentiate w.r.t $x$ this expression you get $$n(x+1)^{n-1}=\sum\_{k=0}^nk\binom nkx^{k-1}$$ Now. substituting $x=1$, you get $$n.2^{n-1}=\sum\_{k=0}^nk\binom nk$$ The first term of he RHS, i.e., the term corresponding to $k=0$ is $0$. Therefore $$\sum\_{k=1}^nk\binom nk=n.2^{n-1}$$I believe this is what you want. You can also have a combinatorial proof for the given expression. I leave that part of the proof for you to think. If you want to prove it through induction, assume that $$\text{'P(n) is true' is implied by} \sum\_{k=1}^nk\binom nk=n.2^{n-1}$$ First we check for the base step $n=1$. $$\sum\_{k=1}^1k\binom 1k=1=1.2^{1-1}$$Hence $P(1)$ is true. Assume that $P(m)$ is true for some $m$. Therefore $$\sum\_{k=1}^m k\binom mk=m.2^{m-1}$$ Now we check for $P(m+1)$.$$\begin{align}\sum\_{k=1}^{m+1}k\binom {m+1}k&=\sum\_{k=1}^{m+1}k\left[\binom mk+\binom m{k-1}\right]\\&=\sum\_{k=1}^mk\binom mk+\sum\_{k=0}^m(k+1)\binom mk\\&=2\left[\sum\_{k=0\text{ or }1}^mk\binom mk\right]+\sum\_{k=0}^m\binom mk\\&=2m.2^{m-1}+2^m\\&=(m+1)2^m\end{align}$$ This is the LHS for $P(m+1)$. Therefore $P(m)$ is true $\implies$ $P(m+1)$ is true. Hence proved.