Question

Chứng tỏ rằng \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{^{4^2}}+...+\frac{1}{49^2}+\frac{1}{50^2}

Answer 1

\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{50^2}

Answer 2

\(\frac{1}{2^2}

Answer 3

Đặt tổng sau là B ta có:

\(B=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{49^2}+\frac{1}{50^2}\)

Ta lại có :

\(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};...;\frac{1}{50^2}< \frac{1}{49.50}\)

\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{48.49}+\frac{1}{49.50}\)

\(\Rightarrow B< 1-\frac{1}{50}\)

\(\Rightarrow B< 1\)