Question

[Trần Quỳnh Như]

Tính \(A=1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\)

Answer 1

[Nguyễn Thanh Hằng]

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

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

\(\Leftrightarrow2A-A=\left(2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+......+\dfrac{100}{2^{99}}\right)-\left(1+\dfrac{3}{2^3}+.........+\dfrac{100}{2^{100}}\right)\)

\(\Leftrightarrow A=\dfrac{11}{4}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+......+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

Đặt :

\(H=\dfrac{1}{2^3}+\dfrac{1}{2^4}+......+\dfrac{1}{2^{99}}\)\(\Leftrightarrow A=\dfrac{11}{4}-H-\dfrac{100}{2^{100}}\)

\(\Leftrightarrow2H=\dfrac{1}{2^2}+\dfrac{1}{2^3}+........+\dfrac{1}{2^{98}}\)

\(\Leftrightarrow2H-H=\left(\dfrac{1}{2^2}+\dfrac{1}{2^3}+.....+\dfrac{1}{2^{98}}\right)-\left(\dfrac{1}{2^3}+\dfrac{1}{2^4}+.....+\dfrac{1}{2^{99}}\right)\)

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

\(\Leftrightarrow A=\dfrac{11}{4}+\dfrac{1}{2^2}-\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)