Question

1) Chứng minh: \(\dfrac{1}{n+\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\)

2) Áp dụng tính tổng: A= \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

Giúp mình nha. Mình cảm ơn trước.

Answer 1

1) Ta có :

\(\dfrac{1}{n}-\dfrac{1}{n+1}=\dfrac{n+1}{n\left(n+1\right)}-\dfrac{n}{n\left(n+1\right)}=\dfrac{1}{n\left(n+1\right)}\)

Vậy \(\dfrac{1}{n\left(n+1\right)}=\dfrac{1}{n}-\dfrac{1}{n+1}\rightarrowđpcm\)

2) \(A=\dfrac{1}{1.2}+\dfrac{1}{2.3}+............+\dfrac{1}{99.100}\)

\(\Leftrightarrow A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+......+\dfrac{1}{99}-\dfrac{1}{100}\)

\(\Leftrightarrow A=1-\dfrac{1}{100}\)

\(\Leftrightarrow A=\dfrac{99}{100}\)