Question

[Lưu Nguyễn Quỳnh Trang]

B=1+2+2²+2³+.....+2¹¹

Chứng tỏ rằng B chia hết cho 7

Answer 1

[Nguyễn Nam]

\(B=1+2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^{10}+2^{11}\)

\(\Rightarrow B=\left(1+2+2^2\right)+\left(2^3+2^4+2^5\right)+\left(2^6+2^7+2^8\right)+\left(2^9+2^{10}+2^{11}\right)\)

\(\Rightarrow B=\left(1+2+2^2\right)+2^3\left(1+2+2^2\right)+2^6\left(1+2+2^2\right)+2^9\left(1+2+2^2\right)\)

\(\Rightarrow B=7+2^3.7+2^6.7+2^9.7\)

\(\Rightarrow B=7\left(1+2^3+2^6+2^9\right)⋮7\)

Vậy \(B⋮7\)

Answer 2

[Lưu Nguyễn Quỳnh Trang]

cảm ơn bạn nhiều lắm nha!!!!