Question

Tính

A=1+2+3+4+5+.....+99+100

B= \(\dfrac{1}{2}\)+ \(\dfrac{1}{6}\)+\(\dfrac{1}{12}\)+\(\dfrac{1}{20}\)+ \(\dfrac{1}{30}\)+....+\(\dfrac{1}{9900}\)

Answer 1

A=1+2+3+4+5+...+99+100

A=\(\dfrac{100.\left(100+1\right)}{2}\)=5050

Vậy A=5050

B=\(\dfrac{1}{2}+\dfrac{1}{6}+\dfrac{1}{12}+...+\dfrac{1}{9900}\)

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

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

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

Vậy B=\(\dfrac{99}{100}\)