Question

[phạm ngọc thái]

So sánh

A=\(\frac{1}{2}\)+\(\frac{1}{2^2}\)+...+\(\frac{1}{2^{50}}\)với 1

Answer 1

[phạm ngọc thái]

Toán nha mình nhầm

Answer 2

[Nguyễn Phương Uyên]

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\)

\(2A-A=\left(1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{49}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{50}}\right)\)

\(A=1-\frac{1}{2^{50}}< 1\)