Question

[Bùi Thảo Ly]

1.Chứng tỏ

A=1+2+2^2+2^3+...+2^20 chia hết cho 3

B=1+2+2^2+2^3+...+2^21 ko chia hết cho 3

Answer 1

[Nguyệt]

dậy sớm thế =)

\(A=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^{19}+2^{20}\right)\)

\(A=3+2^2.\left(1+2\right)+...+2^{19}.\left(1+2\right)\)

\(A=3+2^2.3+...+2^{19}.3\)

\(A=3.\left(1+2^2+...+2^{19}\right)\)

Answer 2

[Nguyệt]

\(B=1+\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{20}+2^{21}\right)\)

\(B=1+2.\left(1+2\right)+2^3.\left(1+2\right)+...+2^{20}.\left(1+2\right)\)

\(B=1+2.3+2^3.3+...+2^{20}.3\)

\(B=1+3.\left(2+2^3+...+2^{20}\right)\)

vì \(3.\left(2+2^3+...+2^{20}\right)⋮3,1⋮̸3=>1+3.\left(2+2^3+...+2^{20}\right)⋮̸3\)

câu A mk quên vt chia hết cho 3 bn them vào tí là đc :>