Question

Bài1: Tìm x biết:

2/1.3+2/3.5+2/5.7+.......+2/x(x+2)=2015/2016

Bài2: Chứng minh

a) S=1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/2

b) S=1/2+1/2²+1/2³+........+1/2²⁰<1

Answer 1

Bài 2

a) Ta có

S = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\)

S = \(\dfrac{1}{5}+\left(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\right)\)

Vì \(\dfrac{1}{13}< \dfrac{1}{12}\)

\(\dfrac{1}{14}< \dfrac{1}{12}\)

\(\dfrac{1}{15}< \dfrac{1}{12}\)

=> \(\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}< \dfrac{1}{12}.3\)

Lại có

\(\dfrac{1}{61}< \dfrac{1}{60}\)

\(\dfrac{1}{62}< \dfrac{1}{60}\)

\(\dfrac{1}{63}< \dfrac{1}{60}\)

=> \(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}< \dfrac{1}{60}.3\)

=> S = \(\dfrac{1}{5}+\dfrac{1}{13}+\dfrac{1}{14}+\dfrac{1}{15}+\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}\) < \(\dfrac{1}{5}+\dfrac{1}{12}.3+\dfrac{1}{60}.3\)

= \(\dfrac{1}{5}+\dfrac{1}{4}+\dfrac{1}{20}\) = \(\dfrac{1}{2}\)

=> đpcm

Answer 2

Ta có

\(\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{x\left(x+2\right)}=\dfrac{2015}{2016}\)

\(\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{x}-\dfrac{1}{x+2}=\dfrac{2015}{2016}\)

\(\dfrac{1}{1}-\dfrac{1}{x+2}=\dfrac{2015}{2016}\)

\(\dfrac{1}{x+2}=\dfrac{1}{1}-\dfrac{2015}{2016}\)

\(\dfrac{1}{x+2}=\dfrac{1}{2016}\)

2016 = x + 2

x = 2016 - 2

x = 2014

Vậy x = 2014 là giá trị cần tìm

Answer 3

b) Ta có

S = 1/2 + 1/2² + 1/2³ + ... + 1/2²⁰

2S = 1 + 1/2 + 1/2² + ... + 1/2¹⁹

=> 2S - S = 1 - 1/2²⁰

S = 1 - 1/2²⁰ < 1

=> Ta có đpcm