E=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{98.99.100}\)
* Áp dụng công thức: \(\dfrac{k}{n.\left(n+k\right)}\)=\(\dfrac{1}{n}-\dfrac{1}{n+k}\)
ta có : \(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-....+\dfrac{1}{98.99}-\dfrac{1}{99.100}\)
E=\(\dfrac{1}{1.2}-\dfrac{1}{99.100}\)
E= ........(tính ra)
Giải:
\(E=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}.\)
Áp dung tính chất:
\(\dfrac{2m}{b\left(b+1\right)\left(b+2\right)}=\dfrac{1}{b\left(b+1\right)}-\dfrac{1}{\left(b+m\right)\left(b+2\right)}\), ta có:
\(2E=2\left(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{98.99.100}\right).\)
\(2E=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{98.99.100}.\)
\(2E=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{98.99}-\dfrac{1}{99.100}.\)
\(2E=\dfrac{1}{1.2}+\left(\dfrac{1}{3}-\dfrac{1}{3}\right)+\left(\dfrac{1}{4}-\dfrac{1}{4}\right)+...+\left(\dfrac{1}{98.99}-\dfrac{1}{98.99}\right)-\dfrac{1}{99.100}.\)
\(2E=\dfrac{1}{1.2}+0+0+...+0-\dfrac{1}{99.100}.\)
\(2E=\dfrac{1}{1.2}-\dfrac{1}{99.100}.\)
\(2E=\dfrac{1}{2}-\dfrac{1}{9900}.\)
\(2E=\dfrac{4950}{9900}-\dfrac{1}{9900}.\)
\(2E=\dfrac{4949}{9900}.\)
\(\Rightarrow E=\dfrac{4949}{9900}:2.\)
\(\Rightarrow E=\dfrac{4949}{9900}.\dfrac{1}{2}=\dfrac{4949}{19800}.\)
Vậy \(E=\dfrac{4949}{19800}.\)
~ Học tốt!!! ~
