Question

[thi hue nguyen]

Chứng tỏ 2 + 2² + 2³ + ... + 2¹⁰⁰  chia hết cho 5 và 31

Answer 1

[Không Tên]

\(A=2+2^2+2^3+...+2^{100}\)

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

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

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

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

\(A=2+2^2+2^3+2^4+...+2^{100}\)

\(=\left(2+2^2+2^3+2^4+2^5\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(=2\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)

\(=\left(1+2+2^2+2^3+2^4\right)\left(2+...+2^{96}\right)\)

\(=31\left(2+...+2^{96}\right)⋮31\)

Answer 2

[I don't need you]

ta có: 2 + 2² +2³ + ...+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)

= 2.15 + 2⁵.15 + ...+ 2⁹⁷.15

= 15.(2+2⁵+...+2⁹⁷) chia hết cho 5 ( 15 chia hết cho 5)

=> đ p c m

ta có: 2 +2²+2³+...+2¹⁰⁰

= (2+2²+2³+2⁴+2⁵) + (2⁶+2⁷+2⁸+2⁹+2¹⁰) + ...+ (2⁹⁶+2⁹⁷+2⁹⁸+2⁹⁹+2¹⁰⁰)

= 2.(1+2+2²+2³+2⁴) + 2⁶.(1+2+2²+2³+2⁴) + ...+ 2⁹⁶.(1+2+2²+2³+2⁴)

= 2.31 + 2⁶.31 + ...+2⁹⁶.31

= 31.(2+2⁶+...+2⁹⁶) chia hết cho 31

=> đ p cm