Question

\(\frac{2^2}{1.3}+\frac{3^2}{2.4} +\frac{4^2}{3.5}+...+\frac{100^2}{99.101}\)
Giúp mình với.

HELP ME!

Answer 1

mk ko ghi đb nhé

\(=\frac{1\cdot3+1}{1\cdot3}+\frac{2\cdot4+1}{2\cdot4}+...+\frac{99\cdot101+1}{99\cdot101}.\)

\(=1+\frac{1}{1\cdot3}+1+\frac{1}{2\cdot4}+...+1+\frac{1}{99\cdot101}\)

\(=99+\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{2\cdot4}+...+\frac{2}{99\cdot101}\right)\)

\(=99+\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{2}-\frac{1}{4}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{98}-\frac{1}{100}+\frac{1}{99}-\frac{1}{101}\right)\)

\(=99+\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{100}-\frac{1}{101}\right)\)

phần còn lại bn tự tính nha

Answer 2

chúc lần nữa ngủ ngon nha.<3