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![](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)
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)
![](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)
\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 4
b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(S=5+5^2+...+5^{2006}\)
\(5S=5^2+5^3+...+5^{2007}\)
\(5S-S=5^2+5^3+...+5^{2007}-5-5^2-...-5^{2006}\)
\(4S=5^{2007}-5\)
\(S=\dfrac{5^{2007}-5}{4}\)
b) Ta có:
\(S=5+5^2+...+5^{2006}\)
\(S=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{2005}+5^{2006}\right)\)
\(S=\left(5+25\right)+5^2\cdot\left(5+25\right)+...+5^{2004}\cdot\left(5+25\right)\)
\(S=30+5^2\cdot30+...+5^{2004}\cdot30\)
\(S=30\cdot\left(1+5^2+...+5^{2004}\right)\)
Vậy: S ⋮ 30
![](https://rs.olm.vn/images/avt/0.png?1311)
\(S=1+3+3^2+3^3+...+3^8+3^9\)
\(=1+3+3^2\left(1+3\right)+...+3^8\left(1+3\right)\)
\(=4\left(1+3^2+...+3^8\right)⋮4\)
\(S=\left(1+3\right)+3^2\left(1+3\right)+...+3^8\left(1+3\right)=4\left(1+3^2+...+3^8\right)⋮4\)
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.