Question

[Nguyễn Thị Mỹ Lan]

cho: A = \(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\)so sánh A với 2

Answer 1

[Chỉ_Có_1_Mk_Tôi]

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

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

\(2A-A=\left(1+1+\dfrac{3}{2^2}+...+\dfrac{99}{2^{98}}+\dfrac{100}{2^{99}}\right)-\left(\dfrac{1}{2}+\dfrac{2}{2^2}+\dfrac{3}{2^3}+...+\dfrac{99}{2^{99}}+\dfrac{100}{2^{100}}\right)\)

\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

Đặt:

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

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

\(2B-B=\left(2+1+\dfrac{1}{2^2}+....+\dfrac{1}{2^{97}}+\dfrac{1}{2^{98}}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{98}}+\dfrac{1}{2^{99}}\right)\)

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

Vậy \(A=2-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}< 2\)

Answer 2

[Chỉ_Có_1_Mk_Tôi]

Bây giờ mình đang bận học bài 1 chút.Xíu nữa mình làm cho nhé