Question

B=1+2+2²+2³+.....+2¹¹

Chứng tỏ rằng B chia hết cho 7

Answer 1

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

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

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

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

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

Vậy \(B⋮7\)

Answer 2

mk nghĩ thế này :

B =

B2 = (1+2+2^2 + 2^3 + .... + 2^11) . 2

B2 = 2^1+2^2+2^3+.......+2^12

B2 - B = 2^12 - 1

B = 4096-1 =4095

vi 4095 chia het cho 7 nen b chia het 7

Answer 3

mn k mk va theo doi nha