Proof with a combinations

Question

Prove that $$C^n_1 + 2\times C^n_2 + 3\times C^n_3 + \dots + n \times C^n_n = n\times 2^{n-1}$$ I tried the formula $$C^n_1 + C^n_2 + \dots + C^n_n = 2^n$$ Then , mutiply both sides by $$n$$ $$n\times C^n_1 + n \times C^n_2 +\dots +n\times C^n_n = n\times 2^n$$ I tried to show that the left hand side equals $$2(C^n_1 + 2 \times C^n_2 + \dots + n\times C^n_n )$$ To get the required but i could not get it ?

Answer 1

By Newton Binomial theorem: $$(x+1)^n=\sum_{k=0}^{n}C_{k}^n x^k,$$ so $$n(x+1)^{n-1}=\sum_{k=1}^{n}C_{k}^n k x^{k-1}.$$ Take $x=1$, you will get you equation $$C^n_1 + 2\times C^n_2 + 3\times C^n_3 + \dots + n \times C^n_n = n\times 2^{n-1}.$$

Answer 2

Since $\binom nk=\binom n{n-k}$, we have

$$ \sum_{k=0}^nk\binom nk=\sum_{k=0}^n(n-k)\binom nk\;, $$

and thus

$$ 2\sum_{k=0}^nk\binom nk=\sum_{k=0}^nk\binom nk+\sum_{k=0}^n(n-k)\binom nk=n\sum_{k=0}^n\binom nk=n2^n\;. $$