Question

1/3x1 + 1/3x5 + ......+ 1/95x97 + 1/97x99

Answer 1

\(\frac{1}{1x3}+\frac{1}{3x5}+...+\frac{1}{95x97}+\frac{1}{97x99}\)

\(=\frac{1}{2}x\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+,,,+\frac{1}{97}-\frac{1}{99}\right)\)

\(=\frac{1}{2}x\frac{98}{99}\)

\(=\frac{49}{99}\)

Answer 2

Đặt A=1/3x1 + 1/3x5 + ......+ 1/95x97 + 1/97x99

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

\(2A=1-\frac{1}{99}\)

\(A=\frac{98}{99}:2\)

\(A=\frac{49}{99}\)

Answer 3

sao lại làm như vậy