Question

[Hà Thu Hằng]

tính A=\(\frac{2.1+1}{\left(1.2\right)^2}\)\(+\)\(\frac{2.2+1}{\left(2.3\right)^2}\)\(+\)\(\frac{2.3+1}{\left(3.4\right)^2}\)\(+\)........\(+\)\(\frac{2.99+1}{\left(99.100\right)^2}\)

giúp mk với /thanks

Answer 1

[Nguyễn Ngọc Anh Minh]

\(A=\frac{2^2-1^2}{\left(1.2\right)^2}+\frac{3^2-2^2}{\left(2.3\right)^2}+\frac{4^2-3^2}{\left(3.4\right)^2}+...+\frac{100^2-99^2}{\left(99.100\right)^2}\)

\(A=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+\frac{1}{3^2}-\frac{1}{4^2}+...+\frac{1}{99^2}-\frac{1}{100^2}\)

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