Question

[dân chơi 2 mặt]

CM: P < 1/16 biết P = 1/5^2 + 2/5^2 + 3/5^2 + ... + 11/5^2

Mong mn giúp mik vs ạ

Answer 1

[Xyz OLM]

Sửa đề  P = \(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\)CM P < 5/16

=> 5P = \(1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{11}{5^{10}}\) 

Lấy 5P trừ P theo vế ta có

5P - P = \(\left(1+\frac{2}{5}+\frac{3}{5^2}+...+\frac{11}{5^{10}}\right)-\left(\frac{1}{5}+\frac{2}{5^2}+\frac{3}{5^3}+...+\frac{11}{5^{11}}\right)\)

4P = \(1+\left(\frac{2}{5}-\frac{1}{5}\right)+\left(\frac{3}{5^2}-\frac{2}{5^2}\right)+...+\left(\frac{11}{5^{10}}-\frac{10}{5^{10}}\right)-\frac{11}{5^{11}}\)

4P = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}-\frac{11}{5^{11}}\)

Đặt Q = \(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\)

=> 5Q \(=5+1+\frac{1}{5}+...+\frac{1}{5^9}\)

Lấy 5Q trừ Q theo vế ta có

5Q - Q = \(\left(5+1+\frac{1}{5}+...+\frac{1}{5^9}\right)-\left(1+\frac{1}{5}+\frac{1}{5^2}+...+\frac{1}{5^{10}}\right)\)

4Q \(=5-\frac{1}{5^{10}}\)

=> Q\(=\frac{5}{4}-\frac{1}{5^{10}.4}\)

Khi đó 4P = \(\frac{5}{4}-\frac{1}{5^{10}.4}-\frac{11}{5^{11}}\)

=> P = \(\frac{5}{16}-\frac{1}{5^{10}.16}-\frac{11}{5^{11}.4}\)

\(=\frac{5}{16}-\frac{1}{5^{10}}\left(\frac{1}{16}-\frac{11}{5.4}\right)< \)\(\frac{5}{16}\)

Answer 2

[Nguyễn Minh Đăng]

Bài làm:

Ta có: \(\frac{1}{5^2}+\frac{2}{5^2}+\frac{3}{5^2}+...+\frac{11}{5^2}\)

\(=\frac{1+2+3+...+11}{5^2}=\frac{\left(1+11\right).11:2}{5^2}=\frac{66}{25}>1>\frac{1}{16}\)

\(\Rightarrow P>\frac{1}{16}\)

=> Đề sai