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\(B=3+3^2+3^3+....+3^{120}\)
a, Ta thấy : Cách số hạng của B đều chi hết cho 3
\(B=3+3^2+3^3+....+3^{120}⋮3\)
\(b,B=3+3^2+3^3+....+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+....+\left(3^{119}+3^{120}\right)\)
\(B=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{119}\left(1+3\right)\)
\(B=3.4+3^3.4+...+3^{119}.4\)
\(B=4\left(3+3^3+...+3^{199}\right)\)
Có : \(B=4\left(3+3^3+...+3^{199}\right)⋮4\)
\(\Rightarrow B⋮4\)
\(c,B=3+3^2+3^3+....+3^{120}\)
\(B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{119}+3^{120}\right)\)
\(B=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{118}\left(3+3^2\right)\)
\(B=13+3^2.13+...+3^{118}.13\)
\(B=13\left(3^2+3^4+...+3^{118}\right)\)
Có : \(B=13\left(3^2+3^4+...+3^{118}\right)⋮13\)
\(\Rightarrow B⋮13\)
a)Ta có:S = 2^1 + 2^2 + 2^3 + 2^4 + 2^5 +...+2^199+ 2^200.
=( 2^1 + 2^2) + (2^3 + 2^4) + (2^5+2^6)+...+(2^197+2^198)+(2^199+2^200).
=2.(1+2)+2^3.(1+2)+2^5.(1+2)+...+2^197.(1+2)+2^199(1+2)
=2.3+2^3.3+2^5.3+...+2^197.3+2^199.3
=3.(2+2^3+2^5+...+2^197+2^199)
Vậy tổng S chia hết cho 3.
Xin lỗi bn,mik o làm kịp
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 )
B=(3+32)+(33+34)+...+(359+360)
B=3(1+3)+33(1+3)+34(1+3)+...+359(1+3)
4(4+33+34+...+359)
suy ra:4(4+33+34+...+359)chia hết cho 4
b )
B=(3+32+33)+(34+35+36)+...+(358+359+360)
=3(1+3+9)+34(1+3+9)+...+358(1+3+9)
=13.3+13.34+...+13.358
=13.(3+34+...+358) luôn chia hết cho 13
vậy B chia hết cho 13
a) \(B=3+3^2+3^3+..+3^{60}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)
\(=4\left(3+3^3+...+3^{59}\right)⋮4\)
=>đpcm
b) \(B=3+3^2+3^3+..+3^{60}\)
\(=\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)
\(=3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)=13\left(3+..+3^{58}\right)⋮13\)
=>đpcm
a) \(B=3+3^2+...+3^{90}\)
\(\Leftrightarrow B=\left(3+3^2\right)+...+\left(3^{89}+3^{90}\right)\)
\(\Leftrightarrow B=\left(3+3^2\right)+...+3^{88}.\left(3+3^2\right)\)
\(\Leftrightarrow B=12+...+3^{88}.12\)
\(\Leftrightarrow B=12.\left(1+...+3^{88}\right)⋮4\left(đpcm\right)\)
b)\(B=3+3^2+...+3^{90}\)
\(\Leftrightarrow B=\left(3+3^2\right)+...+\left(3^{89}+3^{90}\right)\)
\(\Leftrightarrow B=\left(3+3^2\right)+...+3^{88}.\left(3+3^2\right)\)
\(\Leftrightarrow B=12+...+3^{88}.12\)
\(\Leftrightarrow B=12.\left(1+...+3^{88}\right)⋮12\left(đpcm\right)\)
c) \(B=3+3^2+...+3^{90}\)
\(\Leftrightarrow B=\left(3+3^2+3^3\right)+...+\left(3^{88}+3^{89}+3^{90}\right)\)
\(\Leftrightarrow B=\left(3+3^2+3^3\right)+...+3^{87}.\left(3+3^2+3^3\right)\)
\(\Leftrightarrow B=39+...+3^{87}.39\)
\(\Leftrightarrow B=39.\left(1+..+3^{87}\right)⋮39\left(đpcm\right)\)
TA CÓ:
A=30+3+32+33+........+311
(30+3+32+33)+....+(38+39+310+311)
3(0+1+3+32)+......+38(0+1+3+32)
3.13+....+38.13 cHIA HẾT CHO 13 NÊN A CHIA HẾT CHO 13( đpcm)
a) \(B=3+3^2+3^3+...+3^{120}\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{199}\left(1+3\right)\)
\(=3.4+3^3.4+3^{199}.4=4\left(3+3^3+...+3^{199}\right)⋮4\)
b) \(B=3+3^2+3^3+...+3^{120}\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{198}\left(1+3+3^2\right)\)
\(=3.13+3^4.13+...+3^{198}.13=13\left(3+3^4+...+3^{198}\right)⋮13\)