# Prove using Newton's Binomial Theorem

Let $n≥1$ be an integer. Prove that $$\sum_{k=0}^n k{n \choose k} = n 2^{n-1}$$

Hint: take the derivative of $(1+x)^n$ . I'm assuming that I need to use Newton's Binomial Theorem here somehow. By Newton's Binomial Theorem $\sum_{k=0}^n {n \choose k} = 2^n$, and derivative of $(1+x)^n$is $n(1+x)^{n-1}$ , if I take $x=1$, I get $n 2^{n-1}$ . I can't understand what's my next step. Any help is highly appreciated.

Use that $$(1 + x)^n = \sum_{k = 0}^n \binom{n}{k}x^k$$

Differentiating both sides with respect to $x$,

$$n(1 + x)^{n-1} = \sum_{k = 0}^n k\binom{n}{k}x^{k-1}$$

Now evaluate both sides at $x = 1$.

• Thank you! This actually makes sense – user2847624 Jan 29 '15 at 21:22

Next derive series term by term:

$$((1+x)^n)'=\left(\sum_{k=0}^{n}{n \choose k}x^k\right)'=\sum_{k=0}^{n}\left({n \choose k}x^k\right)'=\sum_{k=0}^{n}k{n \choose k}x^{k-1}$$

And put $x=1$

You have to take the derivative of $$\sum_{i=0}^{n} \binom{n}{k} x^{k}=(1+x)^{n}$$ and then set x=1 in $$\sum_{i=0}^{n} k\binom{n}{k} x^{k-1}=n(1+x)^{n-1}$$

Let $n$ be a positive integer, and let $$f(x) = (1+x)^n = \sum_{k=0}^n \binom{n}{k}x^k$$ Then $$\frac{df}{dx} = n(1+x)^{n-1}$$ From the first equation RHS, we also have that $$\frac{df}{dx} = \sum_{k=0}^n \binom{n}{k} k x^{k-1}$$ So $$n(1+x)^{n-1} = \sum_{k=0}^n k\binom{n}{k} x^{k-1}$$ Now plug in $x=1$ so that $(1+x) = 2$ and all the $x^{k-1} = 1$ and your relation comes out.