Question

[Đào Thị Hoàng Yến]

Bài 1 .
a) A = 1 + \(\dfrac{1}{2+1}\) + \(\dfrac{1}{2^2+1}\) +\(\dfrac{1}{2^4+1}\) + .....+\(\dfrac{1}{2^{2n}+1}\)

b) B = \(\dfrac{1}{1.2.3}\) + \(\dfrac{1}{2.3.4}\) + \(\dfrac{1}{3.4.5}\) + ... + \(\dfrac{1}{\left(n-1\right).n.\left(n+1\right)}\)

c) C = \(\dfrac{1.2!}{2}\) + \(\dfrac{2.3!}{2^2}\) +... + \(\dfrac{n.\left(n+1\right)!}{2^n}\) (k! = 1. 2 . 3 ... k)

Answer 1

[Ngô Tấn Đạt]

b.

\(B=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+...+\dfrac{1}{\left(n-1\right)n\left(n+1\right)}\\ =\dfrac{1}{2}\left(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+....+\dfrac{2}{\left(n-1\right).n.\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{\left(n-1\right).n}-\dfrac{1}{n\left(n+1\right)}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{n\left(n+1\right)}\right)=\dfrac{1}{4}-\dfrac{1}{2n\left(n+1\right)}\)