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

Tính tổng sau:

$\Large S =C_{n}^{1} 3^{n-1}+2 C_{n}^{2} 3^{n-2}+3 C_{n}^{2} 3^{n-3}+\cdots+n C_{n}^{n}$

• A. chưa có đáp án
• B.
• C.
• D.

Ta có: $\Large S=3^{n} \sum_{k=1}^{n} k C_{n}^{k}\left(\dfrac{1}{3}\right)^{k}$

Vì $\Large k C_{n}^{k}\left(\dfrac{1}{3}\right)^{k}=n\left(\dfrac{1}{3}\right)^{k}=\left(\dfrac{1}{3}\right)^{k} C_{n-1}^{k-1}, \forall k \geq 1$

Nên $\Large S=3^{n} n \sum_{k=1}^{n}\left(\dfrac{1}{3}\right)^{k} C_{n-1}^{k-1}=3^{n-1} n \sum_{k=1}^{n-1}\left(\dfrac{1}{3}\right)^{k} C_{n-1}^{k}$ $\Large =3^{n-1} \cdot n\left(1+\dfrac{1}{3}\right)^{n-1}=n \cdot 4^{n}$