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Lời giải:
$B=3+3^2+(3^3+3^4+3^5)+(3^6+3^7+3^8)+....+(3^{1989}+3^{1990}+3^{1991})$
$=12+3^3(1+3+3^2)+3^6(1+3+3^2)+...+3^{1989}(1+3+3^2)$
$=12+(1+3+3^2)(3^3+3^6+...+3^{1989})$
$=12+13(3^3+3^6+...+3^{1989})$
$\Rightarrow B$ chia $13$ dư $12$.
2/
$B=3+3^2+3^3+...+3^{1991}$
$3B=3^2+3^3+3^4+...+3^{1992}$
$\Rightarrow 3B-B=3^{1992}-3$
$\Rightarrow 2B=3^{1992}-3$
Có:
$3^4\equiv -1\pmod {41}$
$\Rightarrow 3^{1992}=(3^4)^{498}\equiv (-1)^{498}\equiv 1\pmod {41}$
$\Rightarrow 3^{1992}-3\equiv 1-3\equiv -2\pmod {41}$
$\Rightarrow 2B\equiv -2\pmod {41}$
$\Rightarrow 2B\not\vdots 41$
$\Rightarrow B\not\vdots 41$.
Ta có: B= 3 + 33 + 35 + ... + 31991= (3 + 33 + 35) + (37+ 39 + 311 ) + ... + (31987 + 31989 + 31991).
= 3 x (1 + 32 + 34) + 37 x (1 + 32 + 34) + ... + 31987 x (1 + 32 + 34).
= 3 x 91 + 37 x 91 + ... + 31987 x 91= 3 x 7 x 13 + 37 x 7 x 13 + ... + 31987 x 7 x 13.
= 13 x ( 3 x 7 + 37 x 7 + ... + 31987 x 7).
Vì B = 13 x ( 3 x 7 + 37 x 7 + ... + 31987 x 7) nên B chia hết cho 13.
B= (3 + 33 + 35 + 37) + ... + (31985 + 31987 + 31989 + 31991).
= 3 x (1 + 32 + 34 + 36) + ... + 31985 x (1 + 32 + 34 + 36).
= 3 x 820 + ... + 31985 x 820= 3 x 20 x 41 + ... + 31985 x 20 x 41.
= 41 x ( 3 x 20 + .. + 31985 x 20)
Vì B =41 x ( 3 x 20 + .. + 31985 x 20) nên B chia hết cho 41.
b=(3+32 )+(33+34 )+...(31990+31991)
=13+13.33+... + 13. 31990
41 tương tự nhá
1/
$B=3+3^3+3^5+3^7+...+3^{1991}$
$=(3+3^3+3^5)+(3^7+3^9+3^{11})+....+(3^{1987}+3^{1989}+3^{1991})$
$=3(1+3^2+3^4)+3^7(1+3^2+3^4)+...+3^{1987}(1+3^2+3^4)$
$=(1+3^2+3^4)(3+3^7+...+3^{1987})$
$=91(3+3^7+...+3^{1987})=13.7(3+3^7+...+3^{1987})\vdots 13$
2/
$B=(3+3^3+3^5+3^7)+(3^9+3^{11}+3^{13}+3^{15})+....+(3^{1985}+3^{1987}+3^{1989}+3^{1991})$
$=3(1+3^2+3^4+3^6)+3^9(1+3^2+3^4+3^6)+....+3^{1985}(1+3^2+3^4+3^6)$
$=(1+3^2+3^4+3^6)(3+3^9+....+3^{1985})$
$=820(3+3^9+...+3^{1985})$
$=41.20(3+3^9+...+3^{1985})\vdots 41$