Nguyễn Hoàng Giang
10/03/2017 at 18:44
Nguyễn Việt Hoàng 10/03/2017 at 18:53
We have : \(\dfrac{1}{6}+\dfrac{1}{12}+\dfrac{1}{20}+.....+\dfrac{1}{9900}\)
\(=\dfrac{1}{2.3}+\dfrac{1}{3.4}+\dfrac{1}{4.5}+.....+\dfrac{1}{99.100}\)
\(=\dfrac{1}{2}\dfrac{1}{3}+\dfrac{1}{3}\dfrac{1}{4}+.....+\dfrac{1}{99}\dfrac{1}{100}\)
\(=\dfrac{1}{2}\dfrac{1}{100}\)
\(=\dfrac{49}{100}\)

FA FIFA Club World Cup 2018 16/01/2018 at 22:07
We have : 16+112+120+.....+19900
=12.3+13.4+14.5+.....+199.100
=12−13+13−14+.....+199−1100
=12−1100
=49100