how to prove: $\sum\limits_{k=1}^n k\binom{n}{k}=n \cdot 2^{n-1} $ need help to prove this: $\sum\limits_{k=1}^n k\binom{n}{k}=n \cdot 2^{n-1} $
where $n$ is integer $\geq 1$. 

Question also said taking the derivative of $(1 + x)^n$ would be helpful which I've found to be $n(1+x)^{n-1}$

any help appreciated, thank you.
Using the binomial identity shown in $(3)$, we have
$$
\begin{align}
k\binom{n}{k}
&amp;=\binom{k}{1}\binom{n}{k}\\
&amp;=\binom{n}{1}\binom{n-1}{k-1}\tag{1}
\end{align}
$$
Now sum $(1)$:
$$
\begin{align}
\sum_{k=1}^nk\binom{n}{k}
&amp;=\sum_{k=1}^n\binom{n}{1}\binom{n-1}{k-1}\\
&amp;=n2^{n-1}\tag{2}
\end{align}
$$



Binomial Identity
$$
\begin{align}
\binom{n}{k}\binom{k}{j}
&amp;=\lower{10pt}\overset{\displaystyle\overbrace{\binom{n-j}{k-j}\frac{n-j+1}{k-j+1}\cdots\frac{n}{k}}^{\binom{n}{k}}\binom{k}{j}}{\hphantom{\binom{n-j}{k-j}}\underbrace{\hphantom{\frac{n-j+1}{k-j+1}\cdots\frac{n}{k}\binom{k}{j}}}_{\binom{n}{j}}}\\
&amp;=\binom{n-j}{k-j}\binom{n}{j}\tag{3}
\end{align}
$$
They gave you the hint !

Test with $n=4$:

$$(1+x)^4=\binom40x^0+\binom41x^1+\binom42x^2+\binom43x^3+\binom44x^4.$$

After derivation,

$$4\cdot(1+x)^3=0+1\cdot\binom41x^0+2\cdot\binom42x^1+3\cdot\binom43x^2+4\cdot\binom44x^3.$$

Don't you see the connection ?