Question

[Nguyễn Thị Đoan Trang]

Cho \(N=\frac{1.4}{2.3}+\frac{2.5}{3.4}+...+\frac{98.101}{99.100}\). Chứng minh: \(97< N< 98\)

Answer 1

[Aisaka Taiga]

Ta có \(\frac{a\left(a+3\right)}{\left(a+1\right)\left(a+2\right)}=\frac{\left(a+1-1\right)\left(a+2+1\right)}{\left(a+1\right)\left(a+2\right)}=\frac{\left(a+1\right)\left(a+2\right)-\left(a+2\right)+\left(a+1\right)-1}{\left(a+1\right)\left(a+2\right)}\\ \)

= \(1-\frac{2}{\left(a+1\right)\left(a+2\right)}\)

Áp dụng ta có N = \(98-\left(\frac{2}{2.3}+...+\frac{2}{99.100}\right)=98-2.\left(\frac{1}{2.3}+...+\frac{1}{99.100}\right)=98-2.\left(\frac{1}{2}-\frac{1}{100}\right)>97\)