Giá trị biểu thức S= 3^19 C 20 0 + 3^18 C 20 1+ 3^17 C20 2 + ....+ 1/3 C20 20

* Hướng dẫn giải

Đáp án D

Ta có $\frac{{\left(3+x\right)}^{20}}{3}=\frac{{\displaystyle \sum _{k=0}^{20}{C}_{20}^k{3}^{20-k}{x}^k}}{3}=\sum _{k=0}^{20}{C}_{20}^k{3}^{19-k}{x}^k=3^{19}{C}_{20}^0+x^2{3}^{18}{C}_{20}^1+\mathrm{...}+\frac{1}{3}{x}^{20}{C}_{20}^{20}$

Chọn $x=1\Rightarrow \frac{{\left(1+3\right)}^{20}}{3}=3^{19}{C}_{20}^0+3^{18}{C}_{20}^1+\mathrm{...}+\frac{1}{3}{C}_{20}^{20}\Rightarrow S=\frac{4^{20}}{3}$