Question

Chứng minh rằng:

\(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}< \dfrac{3}{4}\)

Answer 1

Đặt A = \(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{100}}\)

\(\Rightarrow3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)

\(2A=3A-A=1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}-\dfrac{100}{3^{100}}\)Đặt B= \(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\)

\(\Rightarrow3B=3+1+\dfrac{1}{3}+...+\dfrac{1}{3^{98}}\)

\(2B=3B-B=3-\dfrac{1}{3^{99}}\)

Nhận xét : 2B < 3 => B < \(\dfrac{3}{2}\)

=> \(B-\dfrac{100}{3^{100}}< \dfrac{3}{2}\) hay 2A < \(\dfrac{3}{2}\)

=> Đpcm

***tik mik nhé***

Answer 2

Ta có :

3M=1+2/3+3/3^2+...+100/3^99

Suy ra :

2M=1+(1/3+1/3^2+1/3^3+...+1/3^99)-100/3^100

Xét B=\(\dfrac{1}{3}+\dfrac{1}{3^2}+\dfrac{1}{3^3}+...+\dfrac{1}{3^{99}}\)

3B=\(1+\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{98}}\)

2B=1-\(\dfrac{1}{3^{99}}\)<1/2

Suy ra : 2M<1+1/2 nên M<3/4