\(\text{Đặt A}=1+\left(2+2^2+2^3+2^4\right)+\left(2^5+2^6+2^7+2^8\right)+...+\left(2^{97}++2^{98}+2^{99}+2^{100}\right)\)
\(=1+2.\left(1+2+2^2+2^3\right)+2^5.\left(1+2+2^2+2^3\right)+...+2^{97}.\left(1+2+2^2+2^3\right)\)
\(=1+2.15+2^5.15+...+2^{97}.15\)
\(=1+15.\left(2+2^5+...+2^{97}\right)\text{ chia 15 dư 1}\)
ta tính ra được 2101 - 1
24 đồng dư vs 1(mod 15)
(24)25 đòng dư vs 125(mod 15)
2100 đồng dư vs 1(mod 15)
2100.2 đồng dư vs 1.2(mod 15)
2101 đồng dư vs 2(mod 15)
2101 -1 đồng dư vs 2-1(mod 15)
2101-1 đồng dư vs 1(mod 15)
chia 15 dư 1(chỗ gạch chân ban viết kí hiệu nhé)
tickkkkkkkkkkkkkkkkk
Ta có
A=2^0+2^1+2^2+2^3+2^4+2^5+2^6+2^7+....+2^97+2^98+2^99+2^100
A=(2^0+2^1+2^2+2^3)+(2^4+2^5+2^6+2^7)+....+(2^97+2^98+2^99+2^100)
A=2^0*(1+2+3+4)+2^4*(1+2+3+4)+.....+2^97*(1+2+3+4)
A=(1+2+4+8)*(2^0+2^4+....+2^97)
A=15*(2^0+2^4+...+2^97)
Vì 15 chia hết cho 15 suy ra 15*(2^0+2^4+..+2^97) cũng phải chia hết cho 15 hay A chia hết cho 15
Vậy A chia 15 dư 0
= 1 + (2+ 22 + 23+ 2 4 ) + ( 25 + 26 + 27 + 28 ) + ............... + ( 297 + 298 + 299 + 2100 )
= 1 + 2 . ( 1+ 22 + 23+ 24) + 25 ( 1+ 22+23+24) + ........... + 297( 1+ 22+23+24)
= 1 + 15. ( 2+ 25 + .....+ 297)
mà 15. ( 2+25+ ......+ 297 ) chia hết cho 15
==> 1 + 15 ( 2+ 25+.....+297 ) chia cho 5 dư 1.