Question

[Lê Nguyên Hùng]

Chưng minh rằng:  (1/2)+(2/2²)+(3/2³)+...+(99/2⁹⁹)+(100/2¹⁰⁰)<2.

Answer 1

[Mr Lazy]

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

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

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

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

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

\(\Rightarrow2B-B=B=2-\frac{1}{2^{99}}\)

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