Question

so sánh\(\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+.......+\frac{1}{n\cdot\left(n+2\right)}\)với\(\frac{1}{2}\)

 

 

Answer 1

Đặt \(A=\frac{1}{3\cdot5}+\frac{1}{5\cdot7}+...+\frac{1}{n\left(n+2\right)}\)

\(2A=\frac{2}{3\cdot5}+\frac{2}{5\cdot7}+...+\frac{2}{n\left(n+2\right)}\)

\(2A=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\)

\(2A=\frac{1}{3}-\frac{1}{n+2}\)

\(2A=\frac{n-1}{3\left(n+2\right)}\)

\(A=\frac{n-1}{6\left(n+2\right)}\)

Ta có : \(\frac{1}{2}=\frac{3\left(n+2\right)}{2\cdot3\left(n+2\right)}=\frac{3n+6}{6\left(n+2\right)}\)

Dễ thấy \(n-1< 3n+6\)

Do đó \(\frac{1}{2}>A\)

Answer 2

1/2×(1/3-1/5+1/5-1/7+.....+1/n-1/n+2)

=> 1/2×(1/3-1/n+2) <1/2

=> 1/3-1/n+2< 1

Vậy 1/3×5+1/5×7+....+1/n×n+2 < 1/2

Answer 3

sai nha bạn

Answer 4

mình gửi trêu chứ tự bt làm

Answer 5

Nguyen Van Hieu             giải thử xem ntn ?

Answer 6

\(\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{n\left(n+2\right)}\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+...+\frac{1}{n}-\frac{1}{n+2}\right)\)

\(=\frac{1}{2}\left(\frac{1}{3}-\frac{1}{n+2}\right)< \frac{1}{2}\forall n\)

Answer 7

Thêm ĐK: \(n\ne0;n\ne-2\) giúp mình!