\(B=3+3^2+3^3+3^4+...+3^{2009}+3^{2010}\)

\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\)

\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)

\(=4.\left(3+3^3+...+3^{2009}\right)\)

⇒ \(B\) ⋮ 4

b: \(C=5\left(1+5+5^2\right)+...+5^{2008}\left(1+5+5^2\right)=31\cdot\left(5+...+5^{2008}\right)⋮31\)

\(A=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\\ A=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(3+...+3^{58}\right)\\ A=13\left(3+...+3^{58}\right)⋮13\)

\(M=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\\ M=\left(2+2^2+2^3+2^4\right)+...+2^{16}\left(2+2^2+2^3+2^4\right)\\ M=\left(2+2^2+2^3+2^4\right)\left(1+...+2^{16}\right)\\ M=30\left(1+...+2^{16}\right)⋮5\)

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.

bạn hình như viết sai đề

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

\(=91\cdot\left(1+...+3^{1998}\right)⋮7\)

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) \(S=5+5^2+5^3+...+5^{100}\)

\(\Rightarrow S=\left(5+5^2\right)+\left(5^3+5^4\right)+...+\left(5^{99}+5^{100}\right)\)

\(\Rightarrow S=\left(5+25\right)+5^2.\left(5+5^2\right)+...+5^{98}.\left(5+5^2\right)\)

\(\Rightarrow S=30+5^2.30+...+5^{98}.30\)

\(\Rightarrow S=\left(1+5^2+...+5^{98}\right).30⋮30\)

\(\Rightarrow S⋮30\left(đpcm\right)\)

khoai vừa S chia hết 31 thím ạ

a)\(S=5+5^2+...+5^{100}\)

\(5S=5\left(5+5^2+...+5^{100}\right)\)

\(5S=5^2+5^3+...+5^{101}\)

\(5S-S=\left(5^2+5^3+...+5^{101}\right)-\left(5+5^2++...+5^{100}\right)\)

\(4S=5^{101}-5\)

\(S=\frac{5^{101}-5}{4}\)

b)Theo câu a ta có:

\(4S+5=5^x\Leftrightarrow5^{101}-5+5=5^x\)

\(\Leftrightarrow5^{101}=5^x\Leftrightarrow x=101\)

c)\(S=5+5^2+...+5^{100}\)

\(=\left(5+5^2+5^3\right)+...+\left(5^{98}+5^{99}+5^{100}\right)\)

\(=5\left(1+5+5^2\right)+...+5^{98}\left(1+5+5^2\right)\)

\(=5\cdot31+...+5^{98}\cdot31\)

\(=31\cdot\left(5+...+5^{98}\right)⋮31\)