Question

Tính S = \(\sqrt{\frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}}+\sqrt{\frac{1}{1^2}+\frac{1}{3^2}+\frac{1}{4^2}}+...+\sqrt{\frac{1}{1^2}+\frac{1}{99^2}+\frac{1}{100^2}}\)

 

Answer 1

Ta có:

\(\sqrt{\frac{1}{1^2}+\frac{1}{n^2}+\frac{1}{\left(n+1\right)^2}}=\sqrt{\frac{n^4+2n^3+3n^2+2n+1}{n^2.\left(n+1\right)^2}}\)

\(=\sqrt{\frac{\left(n^2+n+1\right)^2}{n^2\left(n+1\right)^2}}=\frac{n^2+n+1}{n\left(N+1\right)}=1+\frac{1}{n\left(n+1\right)}\)

\(=1+\frac{1}{n}-\frac{1}{n+1}\)

Thế vào bài toán ta được

\(S=1+1+...+1+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)

\(=98+\frac{1}{2}-\frac{1}{100}=\frac{9849}{100}\)