Question

A= 2+2^2+2^3+...+2^2004. Chứng minh rằng : A chia hết cho 30

Answer 1

2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰

=(2+2²+2³+2⁴)+...(2⁹⁷+2⁹⁸+2⁹⁹+2¹⁰⁰)

=2(1+2+2²+2³)+...+2⁹⁷(1+2+2²+2³)

=2.15+....+2⁹⁷.15

=15(2+...+2⁹⁷)

=> 2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰ chia hết cho 15 (1)

Ta có: 2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰ >2

^=> 2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰ chia hết cho 2 (2)

Từ (1) và (2) => 2+2²+2³+2⁴+...+2⁹⁹+2¹⁰⁰ chia hết cho 30

=> đpcm

Answer 2

A= 2+2^2+2^3+...+2^2004. Chứng minh rằng : A chia hết cho 6

Answer 3

\(2+2^2+2^3+...+2^{2004}\)
\(=\left(2+2^2+2^3+2^4\right)\)\(+\left(2^5+2^6+2^7+2^8\right)\)\(+...+\left(2^{2001}+2^{2002}+2^{2003}+2^{2004}\right)\)
\(=2^0\left(2+2^2+2^3+2^4\right)\)\(+2^4\left(2+2^2+2^3+2^4\right)\)\(+...+2^{2000}\left(2+2^2+2^3+2^4\right)\)
\(=\left(2+2^2+2^3+2^4\right).\)\(\left(1+2^4+2^8+...+2^{2000}\right)\)
\(=30\left(1+2^4+2^8+...+2^{2000}\right)⋮30\)