Question

P=1/5²+1/6²+...+1/100²

Answer 1

de bai la chung minh 1/6<P<1/4

Answer 2

\(P=\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}\)

Đặt tổng A > P

\(A=\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+...+\frac{1}{99.100}\)

\(\Leftrightarrow\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{99.100}>\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(\Leftrightarrow\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{100}>\frac{1}{5^2}+\frac{1}{6^2}+...+\frac{1}{100^2}\)

\(\Leftrightarrow\frac{1}{4}-\frac{1}{100}>P\)

\(\Leftrightarrow\frac{6}{24}>P\Rightarrow P=\frac{6}{25}\)

 

Answer 3

Thêm vào do bạn viết thiếu đề :

\(\frac{6}{25}>\frac{6}{24}\left(\frac{1}{4}\right);\frac{6}{25}>\frac{1}{6}\Rightarrow\frac{1}{6}< \frac{6}{25}< \frac{1}{4}\)

Hay 1/6 < P < 1/4