Câu hỏi của Đỗ Phúc An

Question

A= 1/1.2+...+1/99.100

có công thức thì càng tốt nhe mng!

Toru · Accepted Answer

$A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dots+\dfrac{1}{99\cdot100}$

$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{99}-\dfrac{1}{100}$

$=1-\dfrac{1}{100}=\dfrac{99}{100}$

CT: $\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\left(n\ne0;n\ne-a\right)$

Câu trả lời:

$A=\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+\dfrac{1}{3\cdot4}+\dots+\dfrac{1}{99\cdot100}$

$=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dots+\dfrac{1}{99}-\dfrac{1}{100}$

$=1-\dfrac{1}{100}=\dfrac{99}{100}$

CT: $\dfrac{a}{n\left(n+a\right)}=\dfrac{1}{n}-\dfrac{1}{n+a}\left(n\ne0;n\ne-a\right)$