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b)=3^1+(3^2+3^3+3^4)+(3^5+3^6+3^7)+....+(3^58+3^59+3^60)
=3^1+(3^2.1+3^2.3+3^2.9)+(3^5.1+3^5.3+3^5.9)+......+(3^58.1+3^58.3+3^58.9)
=3^1+3^2.(1+3+9)+3^5.(1+3+9)+.....+3^58.(1+3+9)
=3+3^2.13+3^5.13+.........+3^58.13
=3.13.(3^2+3^5+....+3^58)
vi tich tren co thua so 13 nen tich do chia het cho 13
=
bai1
a) A=(31+32)+(33+34)+...+(359+360)
=(3^1.1+3^1.3)+...+(3^59.1+3^59.2)
=3^1.(1+3)+...+3^59.(1+3)
=3^1.4+....+3^59.4
=4.(3^1+...+3^59)
vi tich tren co thua so 4 nen tich do chia het cho 4
không chia hết cho 12 ghép ba số lại mà tính nhé chúc may mắn
A = 3 + 3^2 + 3^3 + ... + 3^20
A x 3 = 3^2 + 3^3 + 3^4 + ... + 3^21
A x 3 chia hết cho 3 => A chia hết cho 3
\(a,S=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{19}+3^{20}\right)\\ S=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{18}\left(3+3^2\right)\\ S=\left(3+3^2\right)\left(1+3^2+...+3^{18}\right)=12\left(1+3^2+...+3^{18}\right)⋮12\)
\(b,S=\left(3+3^2+3^3+3^4\right)+...+\left(3^{17}+3^{18}+3^{19}+3^{20}\right)\\ S=\left(3+3^2+3^3+3^4\right)+....+3^{16}\left(3+3^2+3^3+3^4\right)\\ S=\left(3+3^2+3^3+3^4\right)\left(1+...+3^{16}\right)\\ S=120\left(1+...+3^{16}\right)⋮120\)
\(a,S=3+3^2+3^3+...+3^{20}\)
Ta thấy:\(3+3^2=12⋮12\)
\(\Rightarrow S=\left(3+3^2\right)+3^2\left(3+3^2\right)+...+3^{18}\left(3+3^2\right)\\ \Rightarrow S=\left(3+3^2\right)\left(1+3^2+...+1^{18}\right)\\ \Rightarrow S=12.\left(1+3^2+...+3^{18}\right)⋮12\\ \left(đpcm\right)\)
\(b,Ta\) \(thấy:\)\(3+3^2+3^3+3^4=120⋮120\)
\(\Rightarrow S=\left(3+3^2+3^3+3^4\right)+...+\left(3^{17}+3^{18}+3^{19}+3^{20}\right)\\ \Rightarrow S=\left(3+3^2+3^3+3^4\right)+...+3^{16}\left(3+3^2+3^3+3^4\right)\\ \Rightarrow S=\left(3+3^2+3^3+3^4\right)\left(1+...+3^{16}\right)\\ \Rightarrow S=120\left(1+...+3^{16}\right)⋮120\\ \left(đpcm\right)\)
A = \(3^1+3^2+3^3+...+3^{60}\)
A = 3 ( 1 + 3 ) + \(3^3\left(1+3\right)\)+ ..... + \(3^{59}\left(1+3\right)\)
A = 3 . 4 + \(3^3.4\) + ..... + \(3^{59}.4\)
A = 4 ( \(3+3^3+....+3^{59}\)) chia hết cho 4
Vậy A = \(3^1+3^2+3^3+...+3^{60}\)chia hết cho 4
a) \(\left(1+2+2^2+...+2^7\right)\)
\(=\left(1+2\right)+\left(2^2+2^3\right)+...+\left(2^6+2^7\right)\)
\(=\left(1+2\right)+2^2.\left(1+2\right)+...+2^6.\left(1+2\right)\)
\(=3+2^2.3+...+2^6.3\)
\(=3.\left(1+2^2+...+2^6\right)⋮3\left(đpcm\right)\)
a) Đặt A = 1 + 2 + 22 + 23 + ... + 27
Ta có:
A = 1 + 2 + 22 + 23 + ... + 27
\(\Rightarrow\)2A = 2 + 22 + 23 + 24 + ... + 28
\(\Rightarrow\)A = 28 - 1 = 255
Vì 255\(⋮\)3\(\Rightarrow\)2 + 22 + 23 + 24 + ... + 28\(⋮\)3
\(\Rightarrow\)ĐPCM
A = 3 + 32 + 33 + 34 + ... 3100
A = 31 + 32 + 33 + 34 + ...... 3100
A = ( 3100 - 31 ) : 11
A = 398 - ( 32 + 34 )
A = 392
A không chia hết cho 12 vì 12 là thừa số nguyên tố chẵn
+) \(A=3+3^2+3^3+3^4+...+3^{100}\)
\(A=3\left(1+3\right)+3^3\left(1+3\right)+....+3^{99}\left(1+3\right)\)
\(\Rightarrow A⋮4\)
+) \(A=3+3^2+3^3+3^4+...+3^{100}\)
\(A=3\left(1+3+3^2\right)+.....\)( tương tự nhóm liên tiếp 3 số )
\(A=3.13+......⋮13\)
\(\Rightarrow A⋮̸12\)
a) \(A=3^1+3^2+3^3+...+3^{60}\)
\(=\left(3^1+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)
\(=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
\(=13\left(3+3^4+...+3^{58}\right)⋮13\)
b) \(B=2+2^2+2^3+...+2^{20}\)
\(=\left(2+2^2+2^3+2^4\right)+...+\left(2^{17}+2^{18}+2^{19}+2^{20}\right)\)
\(=2\left(1+2+2^2+2^3\right)+...+2^{17}\left(1+2+2^2+2^3\right)\)
\(=15\left(2+2^5+...+2^{17}\right)\div5\)