Question

A = 2 2^2 2^3 ... 2^100 chứng minh A là chia hết cho 15

Answer 1

Sửa đề: \(A=2+2^2+2^3+...+2^{100}\)

\(=\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(=2\left(1+2+2^2+2^3\right)+2^5\left(1+2+2^2+2^3\right)+...+2^{97}\left(1+2+2^2+2^3\right)\)

\(=15\left(2+2^5+...+2^{97}\right)⋮15\)

Answer 2

: A=2+22+23+...+2100�=2+22+23+...+2100

=(2+22+23+24)+(25+26+27+28)+...+(297+298+299+2100)=(2+22+23+24)+(25+26+27+28)+...+(297+298+299+2100)

=2(1+2+22+23)+25(1+2+22+23)+...+297(1+2+22+23)=2(1+2+22+23)+25(1+2+22+23)+...+297(1+2+22+23)

=15(2+25+...+297)⋮15

Answer 3

=(2+2\(^2\)+2\(^3\)+2\(^4\))+(2\(^5\)+2\(^6\)+2\(^7\)+2\(^8\))+...+(2\(^{97}\)+2\(^{98}\)+2\(^{99}\)+2\(^{100}\))

=2(1+2+2\(^2\)+2\(^3\))+2\(^5\)(1+2+2\(^2\)+2\(^3\))+...+2\(^{97}\)(1+2+2\(^2\)+2\(^3\))

=15(2+2\(^5\)+...+2\(^{97}\))⋮15