Question

[Thùy's Linh's]

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+......+\dfrac{1}{100.101}\)

Answer 1

[Phạm Tú Uyên]

\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+\dfrac{1}{5.6}+...+\dfrac{1}{100.101}\)\(=\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{100}-\dfrac{1}{101}\)\(=\dfrac{1}{2}-\dfrac{1}{101}=\dfrac{99}{202}\)

Answer 2

[Phạm Thanh Hằng]

CM công thức :

\(\dfrac{1}{n}-\dfrac{1}{n+a}=\dfrac{n+a}{n\left(n+a\right)}-\dfrac{n}{n\left(n+a\right)}=\dfrac{a}{n\left(n+a\right)}\)Nhận xét :

\(\dfrac{1}{2.3}=\dfrac{1}{2}-\dfrac{1}{3};\dfrac{1}{3.4}=\dfrac{1}{3}-\dfrac{1}{4};...;\dfrac{1}{100.101}=\dfrac{1}{100}-\dfrac{1}{101}\)

\(\Rightarrow\)\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{5}+...+\dfrac{1}{100}-\dfrac{1}{101}\)

\(\dfrac{\Rightarrow1}{2}-\dfrac{1}{101}\)

=\(\dfrac{101}{202}-\dfrac{2}{202}=\dfrac{99}{202}\)

~ chúc bn học tốt~