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* B = 3 + 32 + 33 + 34 +...+ 31991
<=> B = ( 3 + 32 + 33 ) + ( 34 + 35 +36 ) +...+ ( 31989 +31990 +31991 )
<=> B = 3( 1 + 3 + 32 ) + 34( 1 + 3 + 32 ) +...+31989( 1 + 3 + 32 )
<=> B = ( 1 + 3 + 32 )( 3 + 34 +...+ 31989 )
<=> B = 13( 3 + 34 +...+ 31989 ) chia hết cho 13
( đpcm )
* B = 3 + 32 + 33 + 34 +...+ 31991
<=> B =
B = 3 + 33 + 35 +...+ 31991
= (3 + 33 + 35 ) + (37 + 39 + 311) +...+ (31987 + 31989 + 31991)
= 3 . (1 + 32 + 34) + 37 . (1 +32 + 34) +...+ 31987 . (1 + 32 +34)
= 3 . 91 +37 . 91 + ...+ 31987 . 91
= 3 . 7. 13 + 37 . 7 .13 +...+ 31987 . 7 .13
= 13 . (3.7 + 37 .7 +...+ 31987.7) \(⋮13\)
B= 3 + 33 +35 +...+ 31991
= ( 3+ 33 + 35 + 37 ) +...+ (31985 + 31987 + 31989 + 31991)
= 3. (1+32 + 34 +36) +...+ 31985 . (1+ 32 +34 +36)
= 3 . 820 +...+ 31985 . 820
= 3 . 20 .41 +...+ 31985 . 20 . 41
= 41. (3.20 +...+ 31985 . 20) \(⋮41\)
A = 2 + 22 + 23 +......+ 260
-> A = ( 2 + 22 ) + ( 23 + 24 ) + ....+ ( 259 + 260 )
-> A = 2.( 1+2 ) + 23.( 1+2) +......+ 259.( 1+2)
-> A = 2.3 + 23.3 +......+ 259.3
-> A= 3.( 2 + 23 +.....+ 259)
Vì 3 chia hết cho 3
-> 3.( 2 + 23 +...+259)
Vậy A chia hết cho 3
A = 2 + 22 + 23 +.......+ 260
-> A = ( 2 + 22 + 23 ) +.......+ ( 258 + 259 + 260 )
-> A = 2.( 1 + 2 + 22 ) +......+ 258 .( 1 + 2 + 22 )
-> A = 2.7 +.....+ 258.7
-> A = 7.( 2 + .....+ 258 )
Vì 7 chia hết cho 7
-> 7.( 2+....+ 258 )
Vậy A chia hết cho 7
A = 2 + 22 + 23 +......+ 260
-> A = ( 2 + 22 + 23 + 24 ) +.....+ ( 257 + 258 + 259 + 260 )
-> A = 2.( 1 + 2 + 22 + 23 ) +.....+ 257.( 1+ 2 + 22 + 23 )
-> A = 2.15 + ......+ 257.15
-> A = 15.( 2 +.... + 257 )
Vì 15 chia hết cho 15
-> 15.( 2 +....+ 257 )
Vậy A chia hết cho 15
Ta có: \(B=3+3^3+3^5+...+3^{1991}\)
\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(=3\left(1+3^2+3^4\right)+3^7\left(1+3^2+3^4\right)+...+3^{1987}\left(1+3^2+3^4\right)\)
\(=91\left(3+3^7+...+3^{1987}\right)⋮13^{\left(đpcm\right)}\)( vì 91 chia hết cho 33)
Phần còn lại chứng minh tương tự
a/ Ta co: \(B=3+3^3+3^5+...+3^{1987}+3^{1989}+3^{1991}\)
\(\Rightarrow B=\left(3+3^3+3^5\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(\Rightarrow B=3\cdot\left(1+3^2+3^4\right)+...+3^{1987}\cdot\left(1+3^2+3^4\right)\)
\(\Rightarrow B=3\cdot91+...+3^{1987}\cdot91\)
\(\Rightarrow B=91\cdot\left(3+...+3^{1987}\right)\)
\(\Rightarrow13\cdot7\cdot\left(3+...+3^{1987}\right)⋮13\left(dpcm\right)\)
a)
\(P=3+3^3+3^5+...+3^{1991}\)
\(P=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(P=\left(3+3^3+3^5\right)+3^6\left(3+3^3+3^5\right)+...+3^{1986}\left(3+3^3+3^5\right)\)
\(P=273+3^6\cdot273+...+3^{1986}\cdot273\)
\(P=13\cdot21+3^6\cdot13\cdot21+...+2^{1986}\cdot13\cdot21\)
\(P=13\left(21+3^6\cdot21+...+3^{1986}\cdot21\right)⋮13\) (đpcm)
b)
\(P=3+3^3+3^5+...+3^{1991}\)
\(P=\left(3+3^5\right)+\left(3^3+3^7\right)+...+\left(3^{1987}\cdot3^{1991}\right)\)
\(P=\left(3+3^5\right)+3^2\left(3+3^5\right)+...+3^{1986}\left(3+3^5\right)\)
\(P=246+3^2\cdot246+...+3^{1986}\cdot246\)
\(P=41\cdot6+3^2\cdot41\cdot6+...+3^{1986}\cdot41\cdot6\)
\(P=41\left(6+3^2\cdot6+...+3^{1986}\cdot6\right)⋮41\) (đpcm)
Vậy ...
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