Tính tổng $\displaystyle S=9+3+1+\frac{1}{3}+\frac{1}{9}+\cdots +\frac{1}{{{3}^{n-3}}}+\cdots $. A. $\displaystyle S=\frac{27}{2}.$ B. $S=14.$ C. $S=16.$ D. $S=15.$
Đáp án đúng: A Ta có $\displaystyle S=9+3+1+\frac{1}{3}+\frac{1}{9}+\cdots +\frac{1}{{{3}^{n-3}}}+\cdots =9\left( \underbrace{1+\frac{1}{3}+\frac{1}{{{3}^{2}}}+\frac{1}{{{3}^{4}}}+\cdots +\frac{1}{{{3}^{n-1}}}+\cdots }_{CSN\,lvh:\,\,{{u}_{1}}=1,\,q=\frac{1}{3}} \right)=9\left( \frac{1}{1-\frac{1}{3}} \right)=\frac{27}{2}.$ Chọn A.