Question

Tính:

A=\(\frac{2.1+1}{\left[1.\left(1+1\right)\right]^2}+\frac{2.2+1}{\left[2.\left(2+1\right)\right]^2}+...+\frac{2.99+1}{\left[99\left(99+1\right)\right]^2}\)

Answer 1

Ta có:

\(\frac{2n+1}{\left[n\left(n+1\right)\right]^2}=\frac{n+n+1}{n^2\left(n+1\right)^2}=\frac{1}{n\left(n+1\right)^2}+\frac{1}{n^2\left(n+1\right)}\)

\(=\frac{1}{n\left(n+1\right)}.\left(\frac{1}{n}+\frac{1}{n+1}\right)=\left(\frac{1}{n}-\frac{1}{n+1}\right).\left(\frac{1}{n}+\frac{1}{n+1}\right)\)

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

Áp dụng vào bài toán ta được

\(A=\frac{2.1+1}{\left[1\left(1+1\right)\right]^2}+\frac{2.2+1}{\left[2\left(2+1\right)\right]^2}+...+\frac{2.99+1}{\left[99\left(99+1\right)\right]^2}\)

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

\(=1-\frac{1}{100^2}=\frac{9999}{10000}\)