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Đặt: \(A=1+3+3^2+3^3+...+3^{1991}\)
\(3A=3+3^2+3^3+...+3^{1992}\)
\(3A-A=3+3^2+3^3+...+3^{1992}-1-3-3^2-...-3^{1991}\)
\(2A=3^{1992}-1\)
\(A=\dfrac{3^{1992}-1}{2}\)
Ta có: B= 3 + 3
3 + 3
5 + ... + 3
1991= ﴾3 + 3
3 + 3
5
﴿ + ﴾3
7+ 3
9 + 3
11
﴿ + ... + ﴾3
1987 + 3
1989 + 3
1991
﴿.
= 3 x ﴾1 + 3
2 + 3
4
﴿ + 3
7 x ﴾1 + 3
2 + 3
4
﴿ + ... + 3
1987 x ﴾1 + 3
2 + 3
4
﴿.
= 3 x 91 + 3
7 x 91 + ... + 3
1987 x 91= 3 x 7 x 13 + 3
7 x 7 x 13 + ... + 3
1987 x 7 x 13.
= 13 x ﴾ 3 x 7 + 3
7 x 7 + ... + 3
1987 x 7﴿.
Vì B = 13 x ﴾ 3 x 7 + 3
7 x 7 + ... + 3
1987 x 7﴿ nên B chia hết cho 13.
B= ﴾3 + 3
3 + 3
5 + 3
7
﴿ + ... + ﴾3
1985 + 3
1987 + 3
1989 + 3
1991
﴿.
= 3 x ﴾1 + 3
2 + 3
4 + 3
6
﴿ + ... + 3
1985 x ﴾1 + 3
2 + 3
4 + 3
6
﴿.
= 3 x 820 + ... + 3
1985 x 820= 3 x 20 x 41 + ... + 3
1985 x 20 x 41.
= 41 x ﴾ 3 x 20 + .. + 3
1985 x 20﴿
Vì B =41 x ﴾ 3 x 20 + .. + 3
1985 x 20﴿ nên B chia hết cho 41.
TK NHA
Ta có: B= 3 + 3 3 + 3 5 + ... + 3 1991= ﴾3 + 3 3 + 3 5 ﴿ + ﴾3 7+ 3 9 + 3 11 ﴿ + ... + ﴾3 1987 + 3 1989 + 3 1991 ﴿. = 3 x ﴾1 + 3 2 + 3 4 ﴿ + 3 7 x ﴾1 + 3 2 + 3 4 ﴿ + ... + 3 1987 x ﴾1 + 3 2 + 3 4 ﴿. = 3 x 91 + 3 7 x 91 + ... + 3 1987 x 91= 3 x 7 x 13 + 3 7 x 7 x 13 + ... + 3 1987 x 7 x 13. = 13 x ﴾ 3 x 7 + 3 7 x 7 + ... + 3 1987 x 7﴿. Vì B = 13 x ﴾ 3 x 7 + 3 7 x 7 + ... + 3 1987 x 7﴿ nên B chia hết cho 13.
B= ﴾3 + 3 3 + 3 5 + 3 7 ﴿ + ... + ﴾3 1985 + 3 1987 + 3 1989 + 3 1991 ﴿. = 3 x ﴾1 + 3 2 + 3 4 + 3 6 ﴿ + ... + 3 1985 x ﴾1 + 3 2 + 3 4 + 3 6 ﴿. = 3 x 820 + ... + 3 1985 x 820= 3 x 20 x 41 + ... + 3 1985 x 20 x 41. = 41 x ﴾ 3 x 20 + .. + 3 1985 x 20﴿ Vì B =41 x ﴾ 3 x 20 + .. + 3 1985 x 20﴿ nên B chia hết cho 41.
a, Chứng minh rằng A chia hết cho 3
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 ) \(⋮3\)
=> đpcm
chứng minh ằng A chia hết cho 7
A = 2+22 + 23 + .....+ 260
A = ( 2+22 + 23 ) + (24 + 25 + 26) + .... + (258+259+260)
A = 2(1+2 +22 ) +24 (1+2 +22 ) + .... +258(1+2 +22 )
A = 2.7 +24.7 + ....+258.7
A= 7(2+24 ....+258 )\(⋮7\)
=> đpcm
Chứng minh A chia hết cho 15
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) \(⋮15\)
=> đpcm
b,
chứng minh chia hết cho 13
B= 3 + 33 + 35 + + ..........+ 31991
B = (3+33 + 35 ) + (37 + 39 +311 ) + ......+ (31987 + 31989 + 31991 )
B = 3(1+32 +34 ) + 37(1+32 + 34 ) + ....+ 31987(1+32 + 34 )
B = 3.91 + 37.91 + ...+ 31987.91
B = 91(3+37 + ... 31987 )
B = 7.13.(3+37 + ... 31987 ) \(⋮13\)
=> đpcm
chứng minh chia hết cho 41
B = 3+33 + 35 + ...+ 31991
B = (3+33 + 35 + 37 ) + ...(31985 + 31987 + 31989 + 31991 )
B = 3(1+32 + 34 + 36 ) + ...+ 31985(1+32 + 34 + 36)
B = 3. 820 + ...+ 31985.820
B = 820(3+...+31985)
B = 20.41 (3+...+31985) \(⋮41\)
=> đpcm
a) A= (2+22)+(23+24)+........(259+260)
= 1(2+22) + 22(2+22) + ....... 258(2+22)
= 1.6 + 22.6 +......... 258.6
=6(1+22+.......258)
Vì 6 chia hết cho 3 nên => 6(1+22+........258)
Các câu còn lại cũng tương tự như vậy nha bn!
B=(3+3^5)+(3^2+3^6)+...+(3^1987+3^1991)
B=3*(1+3^4)+3^2*(1+3^4)+...+3^1987*(1+3^4)
B=3*82+3^2*82+...+3^1987*82
B=82*(3+3^2+...+3^1987)
B=41*2*(3+3^2+...+3^1987)
Nên B chia hết cho 41
Ta có: A= 2 + 22 + 23 + ... + 260= (2 +22) + (23+ 24) + ... + (259 + 260).
= 2 x (2 + 1) + 23 x (2 + 1) + ... + 259 x (2 + 1).
= 2 x 3 + 23 x 3 + ... + 259 x 3.
= 3 x ( 2 + 23 + ... + 259).
Vì A = 3 x ( 2 + 23 + ... + 259) nên A chia hết cho 3.
A= (2 +22 + 23) + (24 + 25 + 26) + ... + (258 + 259 + 260).
= 2 x (1 + 2 + 22) + 24 x (1 + 2 + 22) + ... + 258 x (1 + 2 + 22).
= 2 x 7 + 24 x 7 + ... + 258 x 7.
= 7 x ( 2 + 24 + ... + 258).
Vì A = 7 x ( 2 + 24 + ... + 258) nên A chia hết cho 7.
A= (2 +22 + 23 + 24) + (25 + 26 + 27 + 28) + ... + (257 + 258 + 259 + 260).
= 2 x (1 + 2 + 22 + 23) + 25 x (1 + 2 + 22 + 23) + ... + 257 x (1 + 2 + 22 + 23).
= 2 x 15 + 25 x 15 + ... + 257 x 15.
= 15 x ( 2 + 24 + ... + 258).
Vì A = 15 x ( 2 + 24 + ... + 258) nên A chia hết cho 15.
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.
a) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)
\(=\left(3+3^3+3^5\right)+\left(3^7+3^9+3^{11}\right)+...+\left(3^{1987}+3^{1989}+3^{1991}\right)\)
\(=3\times\left(1+3^2+3^4\right)+3^7\times\left(1+3^2+3^4\right)+...+3^{1987}\times\left(1+3^2+3^4\right)\)
\(=3\times91+3^7\times91+...+3^{1987}\times91\)
\(=3\times7\times13+3^7\times7\times13+...+3^{1987}\times7\times13\)
\(=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)
Vì \(A=13\times\left(3\times7+3^7\times7+...+3^{1987}\times7\right)\)nên A chia hết cho 13.
b) Ta có: \(A=3+3^3+3^5+...+3^{1991}\)
\(=\left(3+3^3+3^5+3^7\right)+...+\left(3^{1985}+3^{1987}+3^{1989}+3^{1991}\right)\)
\(=3\times\left(1+3^2+3^4+3^6\right)+...+3^{1985}\times\left(1+3^2+3^4+3^6\right)\)
\(=3\times820+...+3^{1985}\times820\)
\(=3\times20\times41+...+3^{1985}\times20\times41\)
\(=41\times\left(3\times20+...+3^{1985}\times20\right)\)
Vì \(A=41\times\left(3\times20+...+3^{1985}\times20\right)\)nên A chia hết cho 41.
A = 2 + 22 + ... + 260 chia hết cho 3
=> ( 2 + 22 ) + ( 23 + 24 ) + .... + ( 259 + 260 )
=> 2( 1 + 2 ) + 23( 1 + 2 ) + .... + 259( 1 + 2 )
=> 2 . 3 + 23 . 3 + .... + 259 . 3
=> 3( 2 + ..... + 259 )
=> chia hết cho 3
Những câu khác bạn làm tương tự nhé , tùy vào từng câu mà gộp nhiều hay ít thôi
GOODLUCK !
A={2+2^2}+{2^3+2^4}+.......+{2^59+2^60}
={2.1+2.2}+{2^3.1+2^3.2}+....+{2^59.1+2^59.2}
=2{1+2}+2^3{1+2}+...+2^59{1+2}
=2.3+2^3.3+.....+2^59.3
=3.(2+2^3+...+2^59)
vi co thua so 3 => tich do chia het cho 3
A={2+2^2}+{2^3+2^4}+.......+{2^59+2^60}
={2.1+2.2}+{2^3.1+2^3.2}+....+{2^59.1+2^59.2}
=2{1+2}+2^3{1+2}+...+2^59{1+2}
=2.3+2^3.3+.....+2^59.3
=3.(2+2^3+...+2^59)
vi co thua so 3 => tich do chia het cho 3
B=1+3+\(3^2\)+\(3^3\)+....+\(3^{1991}\)
B=1+3+\(3^2\)+\(3^3\)+....+\(3^{1991}\)
=(1+3+\(3^2\)+\(3^3\))+(\(3^4\)+\(3^5\)+\(3^6\)+\(3^7\))+.....+(\(3^{1988}\)+\(3^{1989}\)+\(3^{1990}\)+\(3^{1991}\))
=(1+\(3^4\))(1+3+\(3^2\)+\(3^3\))(\(3^8\)+....+\(3^{1988}\))
=82.(1+3+\(3^2\)+\(3^3\))(\(3^8\)+....+\(3^{1988}\))
Vì 82⋮41
→E⋮41
→B⋮41(đpcm)
Bạn tham khảo nha:
B=1+3+32+33+....+31991B=1+3+32+33+....+31991
=(1+3+32+33)+(34+35+36+37)+.....+(31988+31989+31990+31991)=(1+3+32+33)+(34+35+36+37)+.....+(31988+31989+31990+31991)
=(1+3+32+33)+34(1+3+32+33)+....+31988(1+3+32+33)=(1+3+32+33)+34(1+3+32+33)+....+31988(1+3+32+33)
=(1+3+32+33)+(1+34+....+31988)=(1+3+32+33)+(1+34+....+31988)
=(1+34)(1+3+32+33)(38+....+31988)=(1+34)(1+3+32+33)(38+....+31988)
=82.(1+3+32+33)(38+....+31988)=82.(1+3+32+33)(38+....+31988)
Vì 82⋮4182⋮41
→82.(1+3+32+33)(38+....+31988)⋮41→82.(1+3+32+33)(38+....+31988)⋮41
→B⋮41(đpcm)
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