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\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
\(=4.\left(3+3^3+...+3^{2009}\right)\)
⇒ \(B\) ⋮ 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)
\(A=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ A=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+...+3^{58}\right)\\ A=13\left(3+...+3^{58}\right)⋮13\)
\(M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\\ M=\left(2+2^2+2^3+2^4\right)+...+2^{16}\left(2+2^2+2^3+2^4\right)\\ M=\left(2+2^2+2^3+2^4\right)\left(1+...+2^{16}\right)\\ M=30\left(1+...+2^{16}\right)⋮5\)
![](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)
a) \(S=5+5^2+5^3+...+5^{100}\)
\(\Rightarrow5S=5^2+5^3+5^4+...+5^{101}\)
\(\Rightarrow5S-S=\left(5^2+5^3+5^4+...+5^{101}\right)-\left(5+5^2+5^3+...+5^{100}\right)\)
\(\Rightarrow4S=5^{101}-5\)
\(\Rightarrow S=\frac{5^{101}-5}{4}\)
b) \(4S+5=5^x\)
\(\Rightarrow5^{101}-5+5=5^x\)
\(\Rightarrow5^{101}=5^x\)
\(\Rightarrow x=101\)
Vậy x = 101
c) \(S=5+5^2+5^3+...+5^{100}\)
\(\Rightarrow S=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)
\(\Rightarrow S=\left(5+25\right)+5^2.\left(5+5^2\right)+...+5^{98}.\left(5+5^2\right)\)
\(\Rightarrow S=30+5^2.30+...+5^{98}.30\)
\(\Rightarrow S=\left(1+5^2+...+5^{98}\right).30⋮30\)
\(\Rightarrow S⋮30\left(đpcm\right)\)
nếu S có thêm 3^0 thì nó không chia hết cho 13 đâu bạn/ đề sai
a) S = 30 + 31 + 32 + .... + 350
31 S= 31 + 32 + .... + 350 + 351
3S - S = 351 - 30 (-) 2S = 351 - 1 (-) S =351 - 1 : 2