Question

Chứng minh M= 1/3 + 1/3^2 + 1/3^3 + 1/3^4 + ....+ 1/3^99 < 1/2

các bn giúp mik với

Answer 1

\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}\right)\)

\(\Rightarrow2M=1-\frac{1}{3^{99}}< 1\Rightarrow M< \frac{1}{2}\left(đpcm\right)\)

Answer 2

đpcm là j bạn

Answer 3

\(M=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+\frac{1}{3^4}+...+\frac{1}{3^{99}}\)

\(\Rightarrow3M=1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\)

\(\Rightarrow3M-M=\left(1+\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

\(\Rightarrow2M=1-\frac{1}{3^{99}}\)

\(\Rightarrow M=\frac{1-\frac{1}{3^{99}}}{2}\)

\(\Rightarrow M< \frac{1}{2}\left(đpcm\right)\)