Question

Tính A

A= 4/3 + 16/15 + 36/35 + .......... + 9604/9603

Answer 1

\(A=1+\frac{1}{3}+1+\frac{1}{15}+...+1+\frac{1}{9603}\)

\(A=1+\frac{1}{1\cdot3}+1+\frac{1}{3\cdot5}+...+1+\frac{1}{97\cdot99}\)

\(A=49+\frac{1}{1\cdot3}+\frac{1}{3\cdot5}+...+\frac{1}{97\cdot99}\)

\(A=49+\frac{1}{2}\left(\frac{2}{1\cdot3}+\frac{2}{3\cdot5}+...+\frac{2}{97\cdot99}\right)\)

\(A=49+\frac{1}{2}\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{97}-\frac{1}{99}\right)\)

\(A=49+\frac{1}{2}\left(1-\frac{1}{99}\right)=49+\frac{49}{99}=\frac{4900}{4851}\)