Question

Cho N =\(\dfrac{1.4}{2.3}+\dfrac{2.5}{3.4}+\dfrac{3.6}{4.5}+...+\dfrac{98.101}{99.100}\)Chứng minh 97<N<98

Answer 1

Ta có 1.4/2.3=(2-1)(3+1)/2.3=1-1/2+1/3-1/2.3

2.5/3.4=(3-1)(4+1)/3.4=1-1/3+1/4-1/3.4

...

Suy ra N=(1-1/2+1/3-1/2.3)+(1-1/3+1/4-1/3.4)+....+(1-1/99+1/100-1/99.100)

N=\(98+\dfrac{1}{100}-\dfrac{1}{2}-\dfrac{1}{2.3}-\dfrac{1}{3.4}-....-\dfrac{1}{99.100}\)

Xét P=\(\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{99.100}\)

P=\(\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+.....+\dfrac{1}{99}-\dfrac{1}{100}\)

P=\(\dfrac{1}{2}-\dfrac{1}{100}\)

Vậy N=98-1+\(\dfrac{1}{50}\)

N=\(97+\dfrac{1}{50}\)

Vậy 97<N<98(ĐPCM)