Question

[Cường Mai]

cmr 1/2+2/2^2+3/2^3+...+100/2^100<2

Answer 1

[I don't need you]

Dat \(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{100}{2^{100}}\)

=> \(2A=1+1+\frac{3}{2^2}+...+\frac{100}{2^{99}}\)

=> \(2A-A=1+1+\frac{3}{2^2}+...+\frac{100}{2^{99}}-\left(\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+...+\frac{100}{2^{100}}\right)\)

=> \(A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)

Dat \(B=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

=> \(2B=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}\)

=> \(2B-B=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\left(1+\frac{1}{2}+...+\frac{1}{2^{99}}\right)\)

=> \(B=2-\frac{1}{2^{99}}\)

=> \(A=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}< 2\)

=> dpcm