Question

[ke giau mat]

cho B=1/2+(1/2)²+(1/2)³+(1/2)⁴+...+(1/2)²⁰¹⁴+(1/2)²⁰¹⁵

Chung minh rang :B<1

Answer 1

[Hoàng Phúc]

\(B=\frac{1}{2}+\left(\frac{1}{2}\right)^2+\left(\frac{1}{2}\right)^3+.....+\left(\frac{1}{2}\right)^{2014}+\left(\frac{1}{2}\right)^{2015}\)

\(B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+.....+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\)

Ta có: \(2B=1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2013}}+\frac{1}{2^{2014}}\)

=>\(2B-B=\left(1+\frac{1}{2}+\frac{1}{2^2}+....+\frac{1}{2^{2013}}+\frac{1}{2^{2014}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2014}}+\frac{1}{2^{2015}}\right)\)

=>\(B=1-\frac{1}{2^{2015}}<1\left(đpcm\right)\)

Answer 2

[Nguyễn Ngọc Anh Minh]

\(2B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2003}}+\frac{1}{2^{2004}}\)

\(B=2B-B=1-\frac{1}{2005}<1\)