Question

M=1/3+1/3^2+1/3^3+....+1/3^99. CMR: M<1/2

Answer 1

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

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

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

Vậy \(M=\left(\frac{1}{3}-\frac{1}{3^{100}}\right):\frac{2}{3}=\frac{1}{2}-\frac{1}{2.3^{99}}<\frac{1}{2}\)

KL: M    < 1/2 (dpcm)