Question

[Trần Thi Hiền]

giá trị của tổng 

A=3/(1*2)^2+5/(2*3)^2+7/(3*4)^2+...+89/(44*45)^2

Answer 1

[Ninh Thế Quang Nhật]

\(A=\frac{3}{\left(1\cdot2\right)^2}+\frac{5}{\left(2\cdot3\right)^2}+\frac{7}{\left(3\cdot4\right)^2}+...+\frac{89}{\left(44\cdot45\right)^2}\)

\(=\frac{2^2-1^2}{1^2\cdot2^2}+\frac{3^2-2^2}{2^2\cdot3^2}+\frac{4^2-3^2}{3^2\cdot4^2}+...+\frac{45^2-44^2}{44^2\cdot45^2}\)

\(=1-\frac{1}{2^2}+\frac{1}{2^2}-\frac{1}{3^2}+....+\frac{1}{44^2}-\frac{1}{45^2}\)

\(=1-\frac{1}{45^2}=1-\frac{1}{2025}=\frac{2024}{2025}\)