\(\frac{2.1+1}{[1.(1+1)]^2} + \frac{2.2+1}{[2.(2+1)]^2} +...+\frac{2.99+1}{[99.(99+1)]^2} \)=A
Gọi k là 1,2,3,...,98,99 ta có \(\frac{2k+1}{\left(k\left(k+1\right)\right)^2}=\frac{1}{k^2}-\frac{1}{\left(k+1\right)^2}\).Từ đó ta có:
\(\frac{2.1+1}{\left(1.\left(1+1\right)\right)^2}=1-\frac{1}{4}\)
\(\frac{2.2+1}{\left(2.\left(2+1\right)\right)^2}=\frac{1}{4}-\frac{1}{9}\)
....
\(\frac{2.99+1}{\left(99.\left(99+1\right)\right)^2}=\frac{1}{99^2}-\frac{1}{100^2}\)
Cộng vế theo vế đc A=\(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{9}+...+\frac{1}{99^2}-\frac{1}{100^2}=1-\frac{1}{100^2}=\frac{100^2-1}{100^2}=\frac{9999}{10000}=0,9999\)
