Tham khảo
Sửa đề:`B=\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}}`
`⇒2B=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}`
`⇒2B-B=\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{99}}-(\frac{1}{2^2}+\frac{1}{2^3}+...+\frac{1}{2^{100}})`
`⇒B=\frac{1}{2}-\frac{1}{2^{100}}`
`⇒B<\frac{1}{2}<1`
Vậy `B<1`