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Ta có: S = 3 + 32 + 33 + 34 + 35 + 36 + 37 + 38 + 39
S = (3 + 32 + 33) + (34 + 35 + 36) + (37 + 38 + 39)
S = 39 + 33(3 + 32 + 33) + 36(3 + 32 + 33)
S = 39 + 33.39 + 36.39
S = 39.(1 + 33 + 36) \(⋮\)-39 (vì 39 \(⋮\)-39)
B = (1 + 3) + (32+33)+.....+(389+390)
= 4 + 32 .(1 + 3) + .....+390.(1+3)
= 1 .4 + 32.4 + ..... +390.4
= 4.(1 + 32 + .... +390) chia hết cho 4
\(S=3+3^2+3^3+3^4+....+3^{89}+3^{90}\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(==3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+3^{88}\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right).\left(3+3^4+....+3^{88}\right)\)
\(=13\left(3+3^4+...+3^{88}\right)\)\(⋮\)\(13\)
Ta có ;
S = 1 + 2 + 2 2 + 2 3 + 2 4 + 2 5 + 2 6 + 2 7
= ( 1 + 2 ) + ( 2 2 + 2 3 ) + ( 2 4 + 2 5 ) + ( 2 6 + 2 7 )
= ( 1 + 2 ) + 2 2 ( 1 + 2 ) + 2 4 ( 1 + 2 ) + 2 6 ( 1 + 2 )
= 3 + 2 2 .3 + 2 4 .3 + 2 6 .3
= 3 . ( 1 + 2 2 + 2 4 + 2 6 ) chia hết cho 3 ( Vì 3 chia hết cho 3 )
A = 3 + 3 2 + 3 3 + ..... + 3 9 + 3 10
= ( 3 + 3 2 ) + ( 3 3 + 3 4 ) .... + ( 3 9 + 3 10 )
= 3 ( 1 + 3 ) + 3 3 . ( 1 + 3 ) + .... + 3 9 ( 1 + 3 )
= 3 . 4 + 3 3 . 4 + .... + 3 9 . 4
= 4 . ( 3 + 33 + ... + 3 9 ) chia hết cho 4 ( Do 4 chia hết cho 4 )
\(S=\left(1+2\right)+\left(2^2+2^3\right)+\left(2^4+2^5\right)+\left(2^6+2^7\right)\)
\(S=3+3\cdot2^2+3\cdot2^4+3\cdot2^6=3\left(1+2^2+2^4+2^6\right)⋮3\)
\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^9+3^{10}\right)\)
\(A=4\cdot3+4\cdot3^3+...+4\cdot3^9=4\cdot\left(3+3^3+...+3^9\right)⋮4\)
Ta có : S = 3 + 32 + 33 + ... + 31998
S = (3 + 32 + 33) + ... + (31996 + 31997 + 31998)
S = 39 + ... + 31995(3 + 32 + 33)
S = 39 + ... + 31995.39
S = 39.(1 + ... + 31995) \(⋮\)39
\(A,\)\(S=\left(3+3^2\right)+\left(3+3^2\right)3^2+...+\left(3+3^2\right)3^{2018} \)
\(\Rightarrow S=9\left(1+3^2+...+3^{2018}\right)\)
\(\Rightarrow S⋮9\)
\(B,\)\(S=3+3^2+3^3+\left(3+3^2+3^3\right)3^3+...\left(3+3^2+3^3\right)3^{2017}\)
\(S=39+39.3^3+...+39.3^{2017}\)
Nhưng xét lại thì thấy 2017 không chia hết cho 3 nên câu b có lẽ sai đề =)))))
\(C,\)\(S=\left(1+3+3^2+3^3\right).3+\left(1+3+3^2+3^3\right).3^4+...+\left(1+3+3^2+3^3\right).3^{2017}\)
\(S=40.3+40.3^4+...+40.3^{2017}\)
\(Vậy...\)
Ta có : \(S=3+3^2+3^3+...+3^9\)
\(=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+\left(3^7+3^8+3^9\right)\)
\(=\left(3+3^2+3^3\right)+3^3\left(3+3^2+3^3\right)+3^6\left(3+3^2+3^3\right)\)
\(=\left(3+3^2+3^3\right)\left(1+3^3+3^6\right)\)
\(=39.\left(1+3^3+3^6\right)⋮\left(-39\right)\) (đpcm)
S = 3 + 32 + 33 + 34 + ..... + 39 . Chia hết cho -39
S = (3 + 32 + 33) + (34 + 35 + 36) + (37 + 38 + 39)
S = 1(3 + 32 + 33) + 33(3 + 32 + 33) + 36(3 + 32 + 33)
S = (1 . 39) + (33 . 39) + (36 . 39)
S = 39 . (1 + 33 + 36) ⋮ (-39)
➤ S ⋮ (-39)