Câu hỏi của Khánh Ly Phan

Question

Tính

B = 1/3 + 1/3³ + 1/3⁵ + ... + 1/3⁹⁹

Nguyễn Minh Đăng · Accepted Answer

Ta có: $B=\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+...+\frac{1}{3^{99}}$

$\Rightarrow9B=3+\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{97}}$

$\Rightarrow9B-B=\left(3+\frac{1}{3}+...+\frac{1}{3^{97}}\right)-\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)$

$\Leftrightarrow8B=3-\frac{1}{3^{99}}$

$\Rightarrow B=\frac{3^{100}-1}{8\cdot3^{99}}$

Câu trả lời:

Ta có: $B=\frac{1}{3}+\frac{1}{3^3}+\frac{1}{3^5}+...+\frac{1}{3^{99}}$

$\Rightarrow9B=3+\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{97}}$

$\Rightarrow9B-B=\left(3+\frac{1}{3}+...+\frac{1}{3^{97}}\right)-\left(\frac{1}{3}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)$

$\Leftrightarrow8B=3-\frac{1}{3^{99}}$

$\Rightarrow B=\frac{3^{100}-1}{8\cdot3^{99}}$