Question

D = \(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\)

Answer 1

\(D=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\)

\(2D=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\right)\)

\(2D=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}\)

\(2D-D=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\right)\)\(D=1-\dfrac{1}{2^{1000}}\)

Answer 2

\(D=\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}.\)

\(2D=2\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\right).\)

\(2D=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}.\)

\(2D-D=\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{999}}\right)-\left(\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+...+\dfrac{1}{2^{1000}}\right).\)

\(D=1+\left(\dfrac{1}{2}-\dfrac{1}{2}\right)+\left(\dfrac{1}{2^2}-\dfrac{1}{2^2}\right)+...+\left(\dfrac{1}{2^{999}}-\dfrac{1}{2^{999}}\right)-\dfrac{1}{2^{1000}.}\)

\(D=1+0+0+...+0-\dfrac{1}{2^{1000}}.\)

\(D=1-\dfrac{1}{2^{1000}}.\)

Vậy.....

Answer 3

2D=\(\dfrac{1}{2^0}+\dfrac{1}{2^1}+.....+\dfrac{1}{2^{999}}\)

2D-D=\(1-\dfrac{1}{2^{999}}\)

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

Answer 4

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