Question

[Bùi Gia Hưng]

so sanh A va B;  A=1/2+1/2^2+1/2^3+.....+1/2^1000; B=1

Answer 1

[Đoàn Đức Hà]

\(A=\frac{1}{2}+\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}\)

\(2A=1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{999}}\)

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

\(A=1-\frac{1}{2^{1000}}< 1=B\)

Answer 2

[Yen Nhi]

`Answer:`

Đặt \(C=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\)

Ta thấy:

\(\frac{1}{1.2}>\frac{1}{2^2}\)

\(\frac{1}{2.3}>\frac{1}{2^3}\)

\(\frac{1}{3.4}>\frac{1}{2^4}\)

...

\(\frac{1}{999.1000}>\frac{1}{2^{1000}}\)

\(\Rightarrow\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{1000}}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{999.1000}\)

\(\Rightarrow A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-\frac{1}{5}+...+\frac{1}{999}-\frac{1}{1000}\)

\(\Rightarrow A< 1-\frac{1}{1000}\)

Mà \(\frac{1}{1000}>0\)

\(\Rightarrow1-\frac{1}{1000}< 1\)

\(\Rightarrow C< B\)

\(\Rightarrow A< C< B\)

\(\Rightarrow A< B\)