Question

[Ngô Hoàng Tùng]

chứng tỏ A=1/2+2/2²+3/2³+...+100/2¹⁰⁰<2

Answer 1

[Nguyệt]

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

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

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

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

\(\Rightarrow A< 1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}\)(3)

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

\(\Rightarrow2P=2+1+\frac{1}{2}+...+\frac{1}{2^{98}}\)

\(\Rightarrow2P-P=2+1+\frac{1}{2}+...+\frac{1}{2^{98}}-1-\frac{1}{2}-...-\frac{1}{2^{99}}\)

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

Từ (1),(2),(3) => A<2

Answer 2

[Tạ Thị Phương Thảo]

Giải

Ta có A =1/2 +  2/2^2 + 3/2^3 + ... + 100/2^100

=> 2A = 1 + 2/2 + 3/2^2 + ... + 100/2^99

=> 2A - A = 1 + 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^99 - 100/2^100

=> A = ( 1 - 100/2^100) + 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^99 (*)

Đặt B = 1/2 + 1/2^2 + 1/2^3 + ... + 1/2^99

=> 2B = 1 + 1/2 + 1/2^2 + ... + 1/2^98

=> 2B - B = 1 - 1/2^99

=> B = 1 - 1/2^99

Thay B vào (*) ta được:

A = ( 1 - 100/2^100 ) + ( 1 - 1/2^99 )

A = 2 - ( 100/2^100 + 1/2^99 ) < 2

=> A < 2 (đpcm)