Question

[Đặng Bình Giang]

Tính

E=\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)có lời giải nhé😘😘😘

Answer 1

[Kuroba Kaito]

E = \(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{99.100}\)

E = \(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{99}-\frac{1}{100}\)

E = \(\frac{1}{2}-\left(\frac{1}{3}-\frac{1}{3}\right)-\left(\frac{1}{4}-\frac{1}{4}\right)-\left(\frac{1}{5}-\frac{1}{5}\right)-...-\left(\frac{1}{99}-\frac{1}{99}\right)-\frac{1}{100}\)

E = \(\frac{1}{2}-\frac{1}{100}\)

E = \(\frac{49}{100}\)