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Ta có :
A= 32+33+34+35+...+350+351
A= (32+33)+(34+35)+...+(350+351)
A= 1(32+33)+32(32+33)+...+348(32+33)
A= 1.36 + 32.36+...+348.36
A= 36(1+32+...+348) \(⋮36\)
Vì A \(⋮36\) mà 36 \(⋮12\)=> A \(⋮12\)
A = (3^2+3^3)+(3^4+3^5)+....+(3^50+3^51)
= 3.(3+3^2)+3^3.(3+3^2)+....+3^49.(3+3^2)
= 3.12 + 3^3.12 + .... +3^49.12
= 12.(3+3^3+....+3^49) chia hết cho 12 (ĐPCM)
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/ \(A=3+3^2+3^3+3^4+.............+3^{49}+3^{50}\)
\(=\left(3+3^2\right)+\left(3^3+3^4\right)+............+\left(3^{49}+3^{50}\right)\)
\(=3\left(1+3\right)+3^3\left(1+3\right)+............+3^{49}\left(1+3\right)\)
\(=3.4+3^3.4+...............+3^{49}.4\)
\(=4\left(3+3^3+...........+3^{49}\right)⋮4\)
\(\Leftrightarrow A⋮4\left(đpcm\right)\)
b/ \(A=3+3^2+3^3+3^4+.............+3^{49}+3^{50}\)
\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^9\right)+........+\left(+3^{47}+3^{48}+3^{49}+3^{50}\right)\)
\(=3\left(1+3+3^2+3^3\right)+3^5\left(1+3+3^2+3^3\right)+........+3^{47}\left(1+3+3^2+3^3\right)\)
\(=3.40+3^5.40+.........+3^{47}.40\)
\(=40\left(3+3^5+...........+3^{47}\right)⋮10\)
\(\Leftrightarrow A⋮10\left(đpcm\right)\)
Bạn lấy 1 và 3, 2 và 4, 5 và 7....48 và 50 cộng với nhau có tổng chia hết cho 10 Suy ra a chia hết cho 10
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)\(A=3+3^2+3^3+3^4+...+3^{49}+3^{50}\)
\(A=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{49}+3^{50}\right)\)
\(A=3.\left(1+3\right)+3^3.\left(1+3\right)+...+3^{49}.\left(1+3\right)\)
\(A=3.4+3^3.4+...+3^{49}.4\)
\(A=4.\left(3+3^3+...+3^{49}\right)⋮4\)
\(\Rightarrow A=3+3^2+3^3+3^4+...+3^{50}⋮4\left(đpcm\right)\)
b) \(A=3+3^2+3^3+3^4+...+3^{49}+3^{50}\)
\(A=\left(3+3^2+3^3+3^4\right)+...+\left(3^{47}+3^{48}+3^{49}+3^{50}\right)\)
\(A=120+...+3^{46}.\left(3+3^2+3^3+3^4\right)\)
\(A=120+...+3^{46}.120\)
\(A=120.\left(1+...+3^{46}\right)⋮10\)
\(\Rightarrow A=3+3^2+3^3+3^4+...+3^{49}+3^{50}⋮10\left(đpcm\right)\)
Vì 13 là lẻ \(\Rightarrow\) 13, 132, 133, 134, 135, 136 là lẻ.
Mà lẻ + lẻ + lẻ + lẻ + lẻ + lẻ = chẵn nên 13 + 132 + 133 + 134 + 135 + 136 là chẵn. \(\Rightarrow\) 13 + 132 + 133 + 134 + 135 + 136 \(⋮\) 2
\(\Rightarrow\) ĐPCM
\(A=1+3+3^2+..........+3^{11}\)
\(\Leftrightarrow A=\left(1+3\right)+\left(3^2+3^3\right)+.........+\left(3^{10}+3^{11}\right)\)
\(\Leftrightarrow A=1\left(1+3\right)+3^2\left(1+3\right)+.........+3^{10}\left(1+3\right)\)
\(\Leftrightarrow A=1.4+3^2.4+.......+3^{10}.4\)
\(\Leftrightarrow A=4\left(1+3^2+..........+3^{10}\right)⋮4\left(đpcm\right)\)
Minh se giup ban
A=3+3^2+3^3+3^4+...+3^50+3^51
A=(3+3^2+3^3)+(3^4+3^5+3^6)+...+(3^49+3^50+3^51)
A=3*(1+3+9)+3^4*(1+3+9)+...+3^49*(1+3+9)
A=3^13+3^4*13+...+3^49*13
Moi thua so cua A deu co thua so 13 nen A chia het cho 13