Question

[NGUYEN TRUNG KIEN]

cm 1/2+2/2^2+3/2^3+4/2^4+...+100/2^100<2

Answer 1

[Duc Loi]

Đặt \(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}\)

\(\Leftrightarrow2A=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{100}{2^{99}}\)

\(\Rightarrow2A-A=A=1+\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)

\(\Leftrightarrow2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}-\frac{100}{2^{99}}\)

\(\Rightarrow2A-A=2-\frac{100}{2^{99}}+\frac{100}{2^{100}}< 2-\frac{100}{2^{100}}+\frac{100}{2^{100}}=2\)

\(\Rightarrow A< 2\Leftrightarrow\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{100}{2^{100}}< 2\left(đpcm\right).\)

Answer 2

[NGUYEN TRUNG KIEN]

cảm ơn nhé