Question

So sánh :

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

Answer 1

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

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

=> \(A=2A-A=1-\frac{1}{2^{50}}< 1\)

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

Answer 2

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

\(\Rightarrow2A=1+\frac{1}{2}+\frac{1}{2^2}+.....+\frac{1}{2^{49}}\)

\(\Rightarrow2A-A=A=1-\frac{1}{2^{50}}< 1\)

Answer 3

Đặt \(S=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{50}}\)

\(\Rightarrow\frac{1}{2}S=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{51}}\)

\(\Rightarrow\frac{1}{2}S-S=-\frac{1}{2}S=\left(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{51}}\right)-\left(\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{50}}\right)\)

\(\Rightarrow-\frac{1}{2}S=\frac{1}{2^{51}}-\frac{1}{2^{50}}\)

ờ tính ra ntn rồi so sánh :)