Question

[le bac hai my]

chứng tỏ rằng:

2 + 2 mũ 2 + 2 mũ 3 + ..............+2 mũ 100 vừa chia hết cho 31, vừa chia hết cho 5

Answer 1

[Thúy Ngân]

Gọi C là giá trị của biểu thức trên

a) CMR : C chia hết cho 31

\(C=2+2^2+2^3+...+2^{99}+2^{100}\)

\(C=\left(2+2^2+2^3+2^4+2^5\right)+\left(2^6+2^7+2^8+2^9+2^{19}\right)+...+\left(2^{96}+2^{97}+2^{98}+2^{99}+2^{100}\right)\)

\(C=2\left(1+2+2^2+2^3+2^4\right)+2^6\left(1+2+2^2+2^3+2^4\right)+...+2^{96}\left(1+2+2^2+2^3+2^4\right)\)

\(C=2.31+2^6.31+...+2^{96}.31\)

\(C=31\left(2+2^6+2^{10}+...+2^{96}\right)⋮31\)(đpcm) 

b) CMR : C chia hết cho 5

\(C=2+2^2+2^3+2^4+...+2^{97}+2^{98}+2^{99}+2^{100}\)

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

\(=2\left(1+2^2\right)+2^2\left(1+2^2\right)+...+2^{97}\left(1+2^2\right)+2^{98}\left(1+2^2\right)\)

=\(2.5+2^2.5+...+2^{97}.5+2^{98}.5\)

\(=5\left(2+2^2+...+2^{97}+2^{98}\right)⋮5\)(đpcm)

Vậy 2 + 2^2 + 2^3 + ...+ 2^98 + 2^99 + 2^100 vừa chia hết cho 5 vừa chia hết cho 31