: Tính tổng sau: $S=C_{n}^{1}{{3}^{n-1}}+2C_{n}^{2}{{3}^{n-2}}+3C_{n}^{3}{{3}^{n-3}}+…+nC_{n}^{n}$

: Tính tổng sau: $S=C_{n}^{1}{{3}^{n-1}}+2C_{n}^{2}{{3}^{n-2}}+3C_{n}^{3}{{3}^{n-3}}+…+nC_{n}^{n}$

C. $n{{.4}^{n-1}}$

B. 0

C. 1

D. ${{4}^{n-1}}$

Hướng dẫn

Chọn A

Ta có: $S={{3}^{n}}\sum\limits_{k=1}^{n}{kC_{n}^{k}{{\left( \frac{1}{3} \right)}^{k}}}$

Vì $kC_{n}^{k}{{\left( \frac{1}{3} \right)}^{k}}=n{{\left( \frac{1}{3} \right)}^{k}}C_{n-1}^{k-1}$ $\forall k\ge 1$nên

$S={{3}^{n}}.n\sum\limits_{k=1}^{n}{{{\left( \frac{1}{3} \right)}^{k}}C_{n-1}^{k-1}}={{3}^{n-1}}.n\sum\limits_{k=0}^{n-1}{{{\left( \frac{1}{3} \right)}^{k}}C_{n-1}^{k}}$$={{3}^{n-1}}.n{{(1+\frac{1}{3})}^{n-1}}=n{{.4}^{n-1}}$.