Question

[Vương Thị Diễm Quỳnh]

bài 18 : tính 

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

Answer 1

[Nguyễn Ngọc Quý]

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

\(2A-A=\left(2-1\right)+\frac{3}{2^2}+\left(\frac{4}{2^3}-\frac{3}{2^3}\right)+\left(\frac{5}{2^4}-\frac{4}{24}\right)+....+\left(\frac{100}{2^{99}}-\frac{99}{2^{99}}\right)-\frac{100}{2^{100}}\)\(A=1+\frac{3}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+...+\frac{1}{2^{99}}-\frac{100}{2^{100}}\)

\(=\left(1+\frac{2}{2^2}\right)+\left(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}\right)-\frac{101}{2^{100}}\)

Đặt B = \(\frac{1}{2^2}+\frac{1}{2^3}+\frac{1}{2^4}+....+\frac{1}{2^{100}}\)

\(2B=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{99}}\)

\(2B-B=\left(\frac{1}{2^2}-\frac{1}{2^2}\right)+.....+\left(\frac{1}{2^{99}}-\frac{1}{2^{99}}\right)+\frac{1}{2}-\frac{1}{2^{100}}=\frac{1}{2}-\frac{1}{2^{100}}\)

\(A=\left(\frac{3}{2}-\frac{101}{2^{100}}\right)+B\)

\(A=\frac{3}{2}-\frac{101}{2^{100}}+\frac{1}{2}-\frac{1}{2^{100}}=\left(\frac{3}{2}+\frac{1}{2}\right)+\left(-\frac{101}{2^{100}}-\frac{1}{2^{100}}\right)\)

\(A=2+\left(-\frac{102}{2^{100}}\right)=2-\frac{102}{2^{100}}\)

Answer 2

[Vương Thị Diễm Quỳnh]

Trang kool Thanh Nguyễn Vinh làm ra đi  chtt ko có