\(S=3^0+3^2+3^4+3^6+........+3^{2002}\)
\(S=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+........+\left(3^{1998}+3^{2000}+3^{2002}\right)\)
\(S=\left(3^0+3^2+3^4\right)+3^6.\left(3^0+3^2+3^4\right)+......+3^{1998}.\left(3^0+3^2+3^4\right)\)
\(S=91+3^6.91+.........+3^{1998}.91\)
\(S=91\left(1+3^6+.........3^{1998}\right)\)
\(S=7.13\left(1+3^6+.......+3^{1998}\right)\)
Mà 7 \(⋮\)7 \(\Rightarrow S⋮7\)
Vậy S chia hết cho 7 ( đpcm )
P.S: Điều phải chứng minh
S= 3^0 +3^2 +3^4 +....+ 3^2002
9S= 3^4 +3^6+.......+3^2004
9S-S=3^2004-1
8S=3^2004-1
S=3^2004-1/8
Nhớ k mik nha!
S=(3^0+3^4+3^6).1+(3^0+3^4+3^6).3^8+3^6+3^6...(3^0+3^4+3^6).3^2002+3^1996+3^1996
S=(3^0+3^4+3^6).(1+3^8+3^6+3^6...3^2002+3^1996+3^1996)
S=91.(1+3^8+3^6+3^6...3^2002+3^1996+3^1996)
Vì 91 chia hết cho 7 nên S chia hết cho 7
Đúng đó bạn 100 %
