\(S=3^0+3^2+3^4+3^6+...+3^{2014}\)

\(=1+3^2+3^4+3^6+...+3^{2014}\)

\(=\left(1+3^2\right)+3^4\left(1+3^2\right)+...+3^{2012}\left(1+3^2\right)\)

\(=7+3^4.7+...+3^{2012}.7=7\left(1+3^4+...+3^{2012}\right)⋮7\)

Vậy ta có đpcm 

Nhân S với 3^2 ta được 9S=3^2+3^4+....+3^2002+3^2004
=>9S-S=(3^2+3^4+....+3^2004)-(3^0+3^2+....+3^2002)
=>8S=3^2004-1
=>S=(3^2004-1)/8
b,ta có S là sô nguyên nên fải c­­­hung minh 3^2004-1chia hết cho 7
ta có : 3^2004-1=(3^6)^334-1=(3^6-1).M=728.M=7.104.M
=>3^2004 chia hết cho 7. Mặt khác (7;8)=1 nên S chia hết cho 7

b) S=(3⁰+3²+3⁴)+...+(3¹⁹⁹⁸+3²⁰⁰⁰+3²⁰⁰²)

S= 91+...+3¹⁹⁹⁸(1+3²+3⁴)

S=91+...+3¹⁹⁹⁸.91

S=91(1+3⁶+...+3¹⁹⁹⁸)

S=13.7.(1+3⁶+...+3¹⁹⁹⁸) chia hết cho 7

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a, \(S=3^0+3^2+3^4+3^6+...+3^{2002}\)

\(\Rightarrow9S=3^2+3^4+3^6+3^8+...+3^{2004}\)

\(\Rightarrow9S-S=\left(3^2+3^4+3^6+3^8+...+3^{2004}\right)-\left(3^0+3^2+3^4+3^6+...+3^{2002}\right)\)

\(\Rightarrow8S=3^{2004}-1\Rightarrow S=\frac{3^{2004}-1}{8}\)

b, Xét dãy số mũ : 0;2;4;6;...;2002

Số số hạng của dãy số trên là :

( 2002 - 0 ) : 2 + 1 = 1002 ( số )

Ta ghép được số nhóm là :

1002 : 3 = 334 ( nhóm )

Ta có : \(S=\left(3^0+3^2+3^4\right)+\left(3^6+3^8+3^{10}\right)+...+\left(3^{1998}+3^{2000}+3^{2002}\right)\)

\(S=\left(3^0+3^2+3^4\right)+3^6\left(3^0+3^2+3^4\right)+...+3^{1998}\left(3^0+3^2+3^4\right)\)

\(S=1.91+3^6.91+...+3^{1998}.91=\left(1+3^6+...+3^{1998}\right).91\)

Vì : \(91⋮7;1+3^6+...+3^{1998}\in N\Rightarrow S⋮7\) (đpcm)

\(S=3^0+3^2+3^4+3^6+...+3^{2002}\)

\(\Rightarrow3^2S=3^2+3^4+3^6+3^8+...+3^{2002}+3^{2004}\)

\(\Rightarrow9S-S=\left(3^2+3^4+3^6+3^8+...+3^{2002}+3^{2004}\right)-\left(3^0+3^2+3^4+3^6+...3^{2000}+3^{2002}\right)\)

\(\Rightarrow8S=3^{2004}-3^0=3^{2004}-1\)

\(\Rightarrow S=\frac{3^{2004}-1}{8}\)

b) S=(3⁰+3²+3⁴)+...+(3¹⁹⁹⁸+3²⁰⁰⁰+3²⁰⁰²)

S= 91+...+3¹⁹⁹⁸(1+3²+3⁴)

S=91+...+3¹⁹⁹⁸.91

S=91(1+3⁶+...+3¹⁹⁹⁸)

S=13.7.(1+3⁶+...+3¹⁹⁹⁸) chia hết cho 7