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B=(4+4^2+4^3)+....+(4^15+4^16+4^17)
=4.(4^0+4^1+4^2)+....+4^15.(4^0+4^1+4^2)
=4.(1+4+16)+....+4^15.(1+4+16)
=4.21+...+4^15.21
21.(4+...+4^15) chia hết cho 17
Do B : 17
=> B : 17 dư 0.
sao 21.(4+...+4^15) lại chia hết cho 17
bạn giải thik kĩ đc ko
![](https://rs.olm.vn/images/avt/0.png?1311)
Xét \(B=4+4^2+4^3+...+4^{17}\)
\(B=4+\left(4^2+4^3+4^4+4^5\right)+\left(4^6+4^7+4^8+4^9\right)+...+\left(4^{14}+4^{15}+4^{16}+4^{17}\right)\)
\(B=4+4^2\left(1+4+4^2+4^3\right)+4^6\left(1+4+4^2+4^3\right)+...+4^{14}\left(1+4+4^2+4^3\right)\)
\(B=4+4^2\cdot85+4^6\cdot85+...+4^{14}\cdot85\)
\(B=4+85\left(4^2+4^6+...+4^{14}\right)\)
\(B=4+17\cdot5\left(4^2+4^6+...+4^{14}\right)\)
Mà \(17\cdot5\left(4^2+4^6+...+4^{14}\right)⋮17\)
\(\Rightarrow4+17\cdot5\left(4^2+4^6+...+4^{14}\right)\)chia 17 dư 4
Hay \(B\)chia 17 dư 4 (ĐPCM)
\(B=4+4^2+4^3+4^4+........+4^{17}\)
\(B=4+\left(4^2+4^4\right)+\left(4^3+4^5\right)+...+\left(4^{15}+4^{17}\right)\)
\(B=4+4^2\left(1+4^2\right)+.....+4^{15}\left(1+4^2\right)\)
\(B=4+4^2.17+....+4^{15}.17\)
\(B=4+17.\left(4^2+4^3+...+4^{15}\right)\)
\(\Rightarrow\)\(17.\left(4^2+4^3+...+4^{15}\right)\)\(⋮17\)
\(\Rightarrow B:17\)\(dư\)\(4\)
\(\text{Vậy B chia 17 dư 4}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=3+3^2+3^3+...+3^{2009}+3^{2010}=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\)
\(A=3.13+3^4.13+...+3^{2008}.13\)
\(A=13\left(3+3^4+...+3^{2008}\right)\)chia hết cho 13
\(B=\left(4+4^3\right)+\left(4^2+4^4\right)+\left(4^5+4^7\right)+\left(4^6+4^8\right)+...+\left(4^{15}+4^{17}\right)\)
\(B=4.17+4^2.17+4^5.17+...+4^{15}.17\)chia hết cho 17=>số dư = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
$A=(4+4^3+4^5+...+4^{17})+(4^2+4^4+4^6+...+4^{16})$
$=[4+(4^3+4^5)+(4^7+4^9)+....+(4^{15}+4^{17})]+[(4^2+4^4)+(4^6+4^8)+...+(4^{14}+4^{16})]$
$=[4+4^3(1+4^2)+4^7(1+4^2)+...+4^{15}(1+4^2)]+[4^2(1+4^2)+4^6(1+4^2)+....+4^{14}(1+4^2)]$
$=4+(1+4^2)(4^3+4^7+...+4^{15}+4^2+4^6+...+4^{14})$
$=4+17(4^3+4^7+...+4^{15}+4^2+4^6+...+4^{14})$
$\Rightarrow A$ chia $17$ dư $4$.
B=(4+4^2+4^3)+....+(4^15+4^16+4^17)
=4.(4^0+4^1+4^2)+....+4^15.(4^0+4^1+4^2)
=4.(1+4+16)+....+4^15.(1+4+16)
=4.21+...+4^15.21
21.(4+...+4^15) chia hết cho 17
vậy B chia hết cho 17