Question

Giúp mình với :

Tính nhanh

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)

Answer 1

Ta có :

\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+..................+\dfrac{1}{99.100}\)

\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.......................+\dfrac{1}{99}-\dfrac{1}{100}\)

\(=1-\dfrac{1}{100}=\dfrac{99}{100}\)

Answer 2

Ta có:

\(\dfrac{1}{1.2}\)+\(\dfrac{1}{2.3}\)+\(\dfrac{1}{3.4}\)+...+\(\dfrac{1}{99.100}\)

= 1- \(\dfrac{1}{2}\)+ \(\dfrac{1}{2}\)-\(\dfrac{1}{3}\)+\(\dfrac{1}{3}\)-\(\dfrac{1}{4}\)+...+\(\dfrac{1}{99}\)-\(\dfrac{1}{100}\)

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

= 1+0+0+...+0+\(\dfrac{-1}{100}\)

= 1+\(\dfrac{-1}{100}\)

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

Answer 3

1/1.2+1/2.3+1/3.4+........1/99.100

Gọi số cần tìm là A

Ta có: A= 1/1-1/2+1/2-1/3+1/3-1/4+........1/99-1/100

A= 1-1/2+1/2-1/3+1/3-1/4+......1/99-1/100

A= 1-1/100

A= 100/100-1/100

Vậy A= 99/100