Prove that $\binom{n}{1}+2\cdot\binom{n}{2}+3\cdot\binom{n}{3}+...+n\cdot\binom{n}{n}=n\cdot2^{n-1}$

Question

(i)$\binom{n}{1}+2\cdot\binom{n}{2}+3\cdot\binom{n}{3}+...+n\cdot\binom{n}{n}=n\cdot2^{n-1}$

(ii)$\binom{n}{1}+2^2\cdot\binom{n}{2}+3^2\cdot\binom{n}{3}+...+n^2\cdot\binom{n}{n}=n(n+1)\cdot2^{n-2}$

I assume that the binomial series can maybe prove the formulas above $(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n$

differentiate the equation and plug in appropriate value/s. I think this question must be quite well known, so even googling might have earned you an answer — user160738, Nov 12 '16 at 19:16
Various other posts about this sum: How to prove this binomial identity $\sum_{r=0}^n {r {n \choose r}} = n2^{n-1}$?, Combinatorial proof of $\sum^{n}_{i=1}\binom{n}{i}i=n2^{n-1}$, Sum of $k {n \choose k}$ is $n2^{n-1}$. — Martin Sleziak, Sep 30 '21 at 14:13

score 4 · Accepted Answer · answered Nov 12 '16 at 19:21

4

$$f_n(x)=(1+x)^n=\binom{n}{0}+\binom{n}{1}x+\binom{n}{2}x^2+...+\binom{n}{n}x^n$$ $$f_n'(x)=n(1+x)^{n-1}=\binom{n}{1}+2x\cdot\binom{n}{2}+3x^2\cdot\binom{n}{3}+...+nx^{n-1}\cdot\binom{n}{n}$$

Evaluate everything for $x=1$ and you have the answer to (i).

answered Nov 12 '16 at 19:21

Alexandre Krajenbrink

539

Prove that $\binom{n}{1}+2\cdot\binom{n}{2}+3\cdot\binom{n}{3}+...+n\cdot\binom{n}{n}=n\cdot2^{n-1}$

1 Answers1