Question

B = \(\frac{1}{2}\)+ \(\frac{1}{2^2}\)+ \(\frac{1}{2^3}\)+ .... + \(\frac{1}{2^{2016}}\). Chứng minh B < 1

Giúp mik nha các bạn . Thanks.  Giải rõ giúp mik nha .

Answer 1

Ta có : 

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

\(2B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\)

\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2015}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}\right)\)

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

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

Vậy \(B< 1\)

Chúc bạn học tốt ~ 

Answer 2

Ta có: 2B=1+1/2+1/2^2+...+1/2^2015

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

B=1-1/2^2015<1

 Vậy B<1

Answer 3

B.2=1+1/2+1/2² +...+1/2²⁰¹⁵

B.2-B=B=1-1/2²⁰¹⁶ <1

=> B<1