Question

Cho \(B=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\)

Chứng tỏ rằng:\(B< 1\)

Answer 1

\(B=\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\)

\(\Rightarrow2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}\)

\(\Rightarrow2B-B=\left(1+\dfrac{1}{2}+...+\dfrac{1}{2^{2015}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2016}}\right)\)

\(\Rightarrow B=1-\dfrac{1}{2^{2016}}< 1\)

\(\Rightarrow B< 1\left(đpcm\right)\)

Vậy...

Answer 2

B= \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{2016}}\)

\(\dfrac{2B=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2015}}}{2B-B=1-\dfrac{1}{2^{2016}}}\)

B= 1- \(\dfrac{1}{2^{2016}}\)<1

=> B<1

Vậy B<1

Answer 3

Ta có : B= 1/2+1/2^2+1/2^3+.....+1/2^2016

suy ra B= (1+1/2+1/2^2+....+1/2^2015)-(1/2+1/2^2+1/2^3+.....+1/2^2016)

B=1-1/2016

B=2015/2016 < 1

Suy ra: B < 1