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I

iceghost

Ta thấy : $\dfrac{1}{n(n+1)}=\dfrac{(n+1)-n}{n(n+1)}=\dfrac{n+1}{n(n+1)}-\dfrac{n}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$
Áp dụng : $1+\dfrac12+\dfrac1{2.3}+\dfrac1{3.4}+\cdots+ \dfrac1{98.99}+\dfrac1{100}$
$=1+\dfrac12+\dfrac12-\dfrac13+\dfrac13-\dfrac14+\cdots+ \dfrac1{98}-\dfrac1{99}+\dfrac1{100} \\
=1+\dfrac12+\dfrac12-\dfrac1{99}+\dfrac1{100} \\
= \cdots$
 
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