Question

Cho S=1/3+3/3.7+5/3.7.11+7/3.7.11.15+...+2n+1/3.7.11...(4n+3),với n thuộc N*.Chứng minh rằng S<1/2

Giúp mình với,mình đang cần gấp

 

Answer 1

Ta có: S = \(\dfrac{1}{3}+\dfrac{3}{3.7}+\dfrac{5}{3.7.11}+...+\dfrac{2n+1}{3.7.11...\left(4n+3\right)}\)

⇒ 2S = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+...+\dfrac{4n+2}{3.7.11...\left(4n+3\right)}\)

⇒ 2S + \(\dfrac{1}{3.7.11...\left(4n+3\right)}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+...+\dfrac{4n+3}{3.7.11...\left(4n+3\right)}\)

Đến đây nó sẽ rút gọn liên tục và sau nhiều lần rút gọn ta có:

2S + \(\dfrac{1}{3.7.11...\left(4n+3\right)}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{10}{3.7.11}+\dfrac{1}{3.7.11}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{11}{3.7.11}\) = \(\dfrac{2}{3}+\dfrac{6}{3.7}+\dfrac{1}{3.7}\) = \(\dfrac{2}{3}+\dfrac{7}{3.7}=\dfrac{2}{3}+\dfrac{1}{3}=1\)

Suy ra 2S < 1 ⇒ S < \(\dfrac{1}{2}\)(đpcm)