Question

Cho;A=\(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+......+\frac{1}{3^{98}}-\frac{1}{3^{100}}\)

Answer 1

CMR: A<0,1

Answer 2

Ta có : \(A=\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+....+\frac{1}{3^{98}}-\frac{1}{3^{100}}\)(1)

=> 3².A = \(1-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^8}+...+\frac{1}{3^{96}}-\frac{1}{3^{98}}\)(2)

Lấy (2) cộng (1) theo vế ta có : 

3².A + A = \(\left(\frac{1}{3^2}-\frac{1}{3^4}+\frac{1}{3^6}-\frac{1}{3^8}+...+\frac{1}{3^{98}}-\frac{1}{3^{100}}\right)+\left(1-\frac{1}{3^2}+\frac{1}{3^4}-\frac{1}{3^6}+...+\frac{1}{3^{96}}-\frac{1}{3^{98}}\right)\)

10A = \(1-\frac{1}{3^{100}}\)

=> A = \(\left(1-\frac{1}{3^{100}}\right):10=\frac{1}{10}-\frac{1}{3^{100}.10}=0,1-\frac{1}{3^{100}.10}< 0,1\)

=> A < 0,1 (ĐPCM)