C=1+31+32+33+...+311
CMR a)C chia hết cho 13 b)C chia hết cho 40
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cho A = 1 + 3 + 32 + 33 + ... + 311
a ) chứng minh A chia hết cho 13
b) chứng minh A chia hết cho 40
A=1+3+3^2+3^3+...+3^98+3^99+3^100
A=(1+3+ 3^2)+(3^3+3^4+3^5)+...+(3^98+3^99+3^100)
A=(1+3+3^2)+3^3x(1+3+3^2)+...+3^98x(1+3+3^2)
A=13x3^3x13+...+3^98x13
=> 13x(1+3+3^3+...+3^98)chia hết cho 13
Vậy A chia hết cho 13
\(C=1+3+3^2+3^3+...+3^{11}\\ a,C=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\left(3^9+3^{10}+3^{11}\right)\\ =13+3^3.\left(1+3+3^2\right)+3^6.\left(1+3+3^2\right)+3^9.\left(1+3+3^2\right)\\ =13+3^3.13+3^6.13+3^9.13\\ =13.\left(1+3^3+3^6+3^9\right)⋮13\)
Ý a phải chia hết cho 13 chứ em?
b: C=(1+3+3^2+3^3)+...+3^8(1+3+3^2+3^3)
=40(1+...+3^8) chia hết cho 40
a: C ko chia hết cho 15 nha bạn
a, C = 1 + 3 1 + 3 2 + 3 3 + . . . + 3 11
= 1 + 3 1 + 3 2 + 3 3 + 3 4 + 3 5 +...+ 3 9 + 3 10 + 3 11
= 1 + 3 1 + 3 2 + 3 3 . 1 + 3 1 + 3 2 + ... + 3 9 1 + 3 1 + 3 2
= 1 + 3 1 + 3 2 . 1 + 3 3 + . . . + 3 9
= 13. 1 + 3 3 + . . . + 3 9 ⋮ 13
b, C = 1 + 3 1 + 3 2 + 3 3 + . . . + 3 11
= 1 + 3 1 + 3 2 + 3 3 + 3 4 + 3 5 + 3 6 + 3 7 + 3 8 + 3 9 + 3 10 + 3 11
= 1 + 3 1 + 3 2 + 3 3 + 3 4 1 + 3 1 + 3 2 + 3 3 + 3 8 1 + 3 1 + 3 2 + 3 3
= 1 + 3 1 + 3 2 + 3 3 . 1 + 3 4 + 3 8
= 40. 1 + 3 4 + 3 8 ⋮ 40
Bài 1:
\(a,A=\left(2+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\\ A=\left(1+2\right)\left(2+2^3+...+2^{2009}\right)=3\left(2+...+2^{2009}\right)⋮3\\ A=\left(2+2^2+2^3\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\\ A=\left(1+2+2^2\right)\left(2+...+2^{2008}\right)=7\left(2+...+2^{2008}\right)⋮7\)
\(b,\left(\text{sửa lại đề}\right)B=\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{2009}+3^{2010}\right)\\ B=\left(1+3\right)\left(3+3^3+...+3^{2009}\right)=4\left(3+3^3+...+3^{2009}\right)⋮4\\ B=\left(3+3^2+3^3\right)+...+\left(3^{2008}+3^{2009}+3^{2010}\right)\\ B=\left(1+3+3^2\right)\left(3+...+3^{2008}\right)=13\left(3+...+3^{2008}\right)⋮13\)
Bài 2:
\(a,\Rightarrow2A=2+2^2+...+2^{2012}\\ \Rightarrow2A-A=2+2^2+...+2^{2012}-1-2-2^2-...-2^{2011}\\ \Rightarrow A=2^{2012}-1>2^{2011}-1=B\\ b,A=\left(2020-1\right)\left(2020+1\right)=2020^2-2020+2020-1=2020^2-1< B\)
a: \(A=\left(1+3\right)+...+3^{10}\left(1+3\right)\)
\(=4\left(1+...+3^{10}\right)⋮4\)
A = 8⁸ + 2²⁰
= (2³)⁸ + 2²⁰
= 2²⁴ + 2²⁰
= 2²⁰.(2⁴ + 1)
= 2²⁰.17 ⋮ 17
Vậy A ⋮ 17
C = 1 + 31 + 32 + 33 + ....... + 311
C = ( 311 - 31 ) : 11 = 39
C = 39 - ( 33 - 32 )
C = 39 - 31
C =38