Question

Cho \(A=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\)
C/m \(A< \frac{1}{2}\)

Answer 1

Ta có :

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

\(\Rightarrow3A-A=\left(1+\frac{1}{3}+....+\frac{1}{3^{98}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+....+\frac{1}{3^{99}}\right)\)

\(\Rightarrow2A=1-\frac{1}{99}\)

\(\Rightarrow A=\frac{1}{2}-\frac{1}{198}< \frac{1}{2}\)

\(\Rightarrow A< \frac{1}{2}\)

Answer 2

=> 3A = 1 + 1/3 + 1/3² +...  +1/3⁹⁸

=> 2A = 1 - 1/3⁹⁹

=> A = \(\frac{1-\frac{1}{3^{99}}}{2}\)

Mà \(1-\frac{1}{3^{99}}\) < 1 nên A < \(\frac{1}{2}\)

Answer 3

Ta có 3A=1/3^2+1/3^3+...+1/3^100

Mà A=1/3+1/3^2+...+1/3^99

suy ra 2A=-1/3+1/3^100

suy ra A=-1/3+1/3^100/2(ghi 1/3+1/3^100 rồi chia cho 2)

A=-1/6.vì -1/6<1/2

suy ra A<1/2(đpcm)

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