Question

[lữ đức lương]

Cho A= 1/2+2/2²+3/2³+4/2⁴+5/2⁵+...+99/2⁹⁹+100/2¹⁰⁰. So sánh A với 2.

Answer 1

[Pham Van Hung]

\(A=\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\frac{4}{2^4}+...+\frac{98}{2^{98}}+\frac{99}{2^{99}}+\frac{100}{2^{100}}\)

\(2A=1+\frac{2}{2}+\frac{3}{2^2}+\frac{4}{2^3}+...+\frac{99}{2^{98}}+\frac{100}{2^{99}}\)

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

Đặt \(B=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{98}}+\frac{1}{2^{99}}\)

\(2B=2+1+\frac{1}{2}+...+\frac{1}{2^{97}}+\frac{1}{2^{98}}\)

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

Do đó: \(A=2-\frac{1}{2^{99}}-\frac{100}{2^{100}}< 2\)