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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
a.
$S=3^0+3^2+3^4+...+3^{2002}$
$3^2S=3^2+3^4+3^6+...+3^{2004}$
$3^2S-S=(3^2+3^4+3^6+...+3^{2004})-(3^0+3^2+3^4+...+3^{2002})$
$8S=3^{2004}-3^0=3^{2004}-1$
$S=\frac{3^{2004}-1}{8}$
b.
$S=(3^0+3^2+3^4)+(3^6+3^8+3^{10})+....+(3^{1998}+3^{2000}+3^{2002})$
$=(3^0+3^2+3^4)+3^6(3^0+3^2+3^4)+....+3^{1998}(3^0+3^2+3^4)$
$=(3^0+3^2+3^4)(1+3^6+...+3^{1998})$
$=91(1+3^6+...+3^{1998})=7.13(1+3^6+...+3^{1998})\vdots 7$
Ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(S=\left(3^0+3^2+3^4\right)+...+3^{1998}\left(3^0+3^2+3^4\right)\)
\(=91\cdot\left(1+...+3^{1998}\right)⋮7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: \(S=3^0+3^2+3^4+...+3^{2002}\)
\(=\left(3^0+3^2+3^4\right)+...+3^{1998}\left(3^0+3^2+3^4\right)\)
\(=91\cdot\left(1+...+3^{1998}\right)⋮7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+...+3^8\right)⋮4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=\left(1+3+3^2\right)+...+3^7\left(1+3+3^2\right)\)
\(=13\left(1+...+3^7\right)⋮13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1+3+3^2+3^3+3^4+3^5+3^6+3^7+3^8+3^9\)
\(S=\left(1+3\right)+\left(3^2+3^3\right)+\left(3^4+3^5\right)+\left(3^6+3^7\right)+\left(3^8+3^9\right)\)
\(S=4+3^2\left(1+3\right)+3^4\left(1+3\right)+3^6\left(1+3\right)+3^8\left(1+3\right)\)
\(S=4+3^2.4+3^4.4+3^6.4+3^8.4\)
\(S=4\left(3^2+3^4+3^6+3^8\right)\)
\(4⋮4\\ \Rightarrow4\left(3^2+3^4+3^6+3^8\right)⋮4\\ \Rightarrow S⋮4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1.\left(1+3\right)+3^2\left(1+3\right)+3^4\left(1+3\right)+...+3^8\left(1+3\right)\)
\(S=4x\left(1+3^2+...+3^8\right)\)
Vì 4 chia hết cho 4 nên S chia hết cho 4
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(S=1+3^2+3^4+3^6+...+3^{98}\)
\(=\left(1+3^2\right)+\left(3^4+3^6\right)+...+\left(3^{96}+3^{98}\right)\)
\(=10+3^4\cdot10+...+3^{96}\cdot10\)
\(=10\left(1+3^4+...+3^{96}\right)⋮10\)(ĐPCM)
S=30+32+34+36+...+32002
9S=32+34+36+38+...+32004
9S-S=32+34+36+...+32004-30+32+34+36+...+32002
8S=32004-30
S=32004-30
8
S = 30 + 32 + 34 + 36 + ... + 32002
S = (30 +32 + 34) + (36 + 38 + 310) + ... + (31998 + 32000 + 32002)
S = 91 + 36.(1+32+34) + ... + 31998.(1+32+34)
S = 91 + 36. 91 + ... + 31998. 91
S= 91. (1+36+...+31998)
S = 7.13. (1+36+...+31998) chia hết cho 7.
Vậy \(S⋮7\).