Giúp bạn bài 1 nhé!
=(2+2^2)+(2^3+2^4)+...+(2^9+2^10)
=2.(1+2)+2^3.(1+2)+...2^9+(1+2)
= 2.3+2^3.3+...+2^9.3 = 3.(2+2^3+2^5+...+2^9) Do 3 chia hết cho 3 Suy ra tổng đó chia hết cho 3
1/ \(A=2+2^2+2^3+...+2^{10}^{ }\)\(=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^9+2^{10}\right)\)\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^9\left(1+2\right)\)\(=2.3+2^3.3+...2^9.3\)\(=3\left(2+2^3+...+2^9\right)\)
Do 3 chia hết cho 3 nên A cũngchia hết cho 3
2/ \(A=4+4^2+...+4^{19}=4+\left(4^2+4^3\right)+...+\left(4^{18}+4^{19}\right)\)\(=4+4^2\left(1+4\right)+4^4\left(1+4\right)+...+4^{18}\left(1+4\right)\)\(=4+4^2.5+4^4.5+...+4^{18}.5\)\(=4+5\left(4^2+4^4+...+4^{18}\right)\)
Mà 5 chia hất cho 5 nên\(5\left(4^2+4^4+...+4^{18}\right)\)cũng chia hết cho 5
Nhưng \(A=4+5\left(4^2+4^4+...+4^{18}\right)\)nên A chia 5 dư 4
A) tong do chia het ch 3
B)so du A: chia het cho 5
chia het cho 4