Question

[nguyen huy anh]

CMR: C=1/3+1/3²+1/3³+...+1/3⁹⁹ nho hon 1/2

Answer 1

[Phạm Hoàng Lan]

mk ko viết lại đề bài nhé !

3C=1+1/3+1/3²+...+1/3⁹⁸

3C-C=2C=1-1/3⁹⁹

C=1/2-1/3⁹⁹.2=>C<1/2 => đpcm

Answer 2

[Xyz OLM]

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

\(\Rightarrow3C=1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\)

\(\text{Lấy }3C-C\text{ ta có : } \)\(3C-C=\left(1+\frac{1}{3}+\frac{1}{3^2}+...+\frac{1}{3^{100}}\right)-\left(\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{99}}\right)\)

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

\(C=\left(1-\frac{1}{3^{99}}\right):2\)

\(C=\left(1-\frac{1}{3^{99}}\right)\times\frac{1}{2}\)

\(C=\frac{1}{2}-\frac{1}{3^{99}\times2}< \frac{1}{2}\)

\(\Rightarrow C< \frac{1}{2}\left(\text{ĐPCM}\right)\)