Question

CHỨNG MINH
  \(\dfrac{1.3+2}{2^{2^{ }}}\)+\(\dfrac{2.4+2}{3^2}\)+\(\dfrac{3.5+2}{4^2}\)+...+\(\dfrac{2008.2010+2}{2009^2}\)+\(\dfrac{2009.2011+2}{2010^2}\) < 2011

GIÚP TỚ ĐI MÀ :))

Answer 1

Trước hết ta chứng minh (a-1)(a+1) + 1 = a^2 (*)

Thật vậy VT = (a-1)(a+1)+1=(a-1)a + a-1 +1 = a^2-a+a=a^2 =VP 

Áp dụng (*) ta có:

\(A=\dfrac{1\cdot3+2}{2^2}+\dfrac{2\cdot4+2}{3^2}+...+\dfrac{2009\cdot2011+2}{2010^2}\\ =\dfrac{2^2+1}{2^2}+\dfrac{3^2+1}{3^2}+...+\dfrac{2010^2+1}{2010^2}=2009+\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{2010^2}\\ < 2009+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{2009\cdot2010}\\ =2009+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+....+\dfrac{1}{2009}-\dfrac{1}{2010}=2010-\dfrac{1}{2010}< 2020< 2011\)