Question

[minh nguyen]

cho A=\(\frac{1}{2^2}+\frac{2}{2^3}+...+\frac{2016}{2^{2017}}\)

chứng minh A<1

Answer 1

[Hà Minh Hiếu]

TA CÓ:

     A = \(\frac{1}{2^2}+\frac{2}{2^3}+...+\frac{2016}{2^{2017}}\)

=> 2A = \(\frac{2.1}{2^2}+\frac{2.2}{2^3}+...+\frac{2016.2}{2^{2017}}\)

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

=> 2A - A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}-\frac{2016}{2^{2017}}\)

=> A = \(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{2016}}-\frac{2016}{2^{2017}}\)

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

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

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

=> A < 1   ( ĐPCM)

Answer 2

[Nguyễn Văn Sơn]