Question

CHỨNG MINH RẰNG

\(Q=\frac{1}{3^1}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}< \frac{1}{2}\)

Answer 1

\(Q=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\)

\(3Q=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\)

\(3Q-Q=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{98}}+\frac{1}{3^{99}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}+\frac{1}{3^{100}}\right)\)

\(2Q=1-\frac{1}{3^{100}}< 1\)

\(\Rightarrow Q=\frac{1-\frac{1}{3^{100}}}{2}< \frac{1}{2}\)

Answer 2

ơ tỉ tỉ toán lớp mấy dợ

Answer 3

Trả lời :......................................................

\(Q=\frac{3^{1-\frac{1}{100}}}{2}< \frac{1}{2}...............................\)

Hk tốt,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,,