Question

1. Tính:

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

Answer 1

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

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

\(2B=1+1+\dfrac{3}{2^2}+......+\dfrac{1000}{2^{999}}\)

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

\(B=1-\dfrac{1000}{2^{999}}\)