Prove $\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$ using binomial and induction

Question

Can anyone help? I've got to prove

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

using binomial first and then induction.

Also https://math.stackexchange.com/questions/2180181/find-the-value-of-sum-k-1nk-binomnk, https://math.stackexchange.com/questions/576089/find-a-formula-for-sum-k-1n-k-n-choose-k, and many more duplicates, which can be found with Approach0 — Martin R, Jan 15 '18 at 09:10

score 0 · Answer 1 · answered Jan 15 '18 at 08:53

0

By binomial theorem we obtain: $$\sum_{i=1}^n\binom{n}{i}i=\left(\sum_{i=1}^n\binom{n}{i}x^i\right)'_{x=1}=\left((1+x)^n\right)'_{x=1}=n(1+x)^{n-1}_{x=1}=n2^{n-1}.$$

answered Jan 15 '18 at 08:53

Michael Rozenberg

203,855

Can you clarify? I'm new to this and I cannot really understand the second term... – user521623 Jan 15 '18 at 09:08
I used $(x^i)'=ix^{i-1}$ – Michael Rozenberg Jan 15 '18 at 09:09
Taking the derivative? – Air Conditioner Jan 15 '18 at 09:17
Yes, of course! – Michael Rozenberg Jan 15 '18 at 09:18

Prove $\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$ using binomial and induction

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