Question

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

Answer 1

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

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

\(2A=2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+....+\dfrac{100}{2^{99}}\)

\(2A-A=\left(2+\dfrac{3}{2^2}+\dfrac{4}{2^3}+\dfrac{5}{2^4}+...+\dfrac{100}{2^{99}}\right)-\left(1+\dfrac{3}{2^3}+\dfrac{4}{2^4}+\dfrac{5}{2^5}+...+\dfrac{100}{2^{100}}\right)\)\(A=2+\dfrac{3}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}+...+\dfrac{1}{2^{99}}-\dfrac{100}{2^{100}}\)

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

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

Answer 2

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

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

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

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

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

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