A=1+3+3^2+3^3+3^4+3^5+3^6+...+3^29.Chứng minh A chia hết cho 13.
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4 tháng 12 2014
A=2^1+2^2+2^3+2^4+...+2^2010
=(2+2^2)+(2^3+2^4)+...+(2^2010+2^2011)
=2.(1+2)+2^3.(1+2)+...+2^2010.(1+2)
=2.3+2^3.3+...+2^2010.3
=(2+2^3+2^2010).3
=> A chia het cho 3
NT
2
S
17 tháng 10 2016
A=3+32+33...+329+330
A=(3+32+33)+...+(328+329+330)
A=3.(1+3+32)+...+328.(1+3+32)
A=3.13+...+328.13
A=13.(3+...+328) chia hết cho 13
17 tháng 10 2016
A= 3(1+3+3^2)+3^4(1+3+3^2)+3^7(1+3+3^2)+...+3^28.(1+3+3^2)
A=(1+3+3^2)(3+3^4+3^7+...+3^25+3^28)
=13.(3+3^4+3^7+...3^28) vậy A chia hết cho 13
2 tháng 10 2016
a) (1+5+52+53+...529)chia hết cho 6
Đặt (1+5+52+53+...529) = A
\(A=\left(1+5\right)+\left(5^2+5^3\right)+\left(5^4+5^5\right)....+\left(5^{28}+5^{29}\right)\)
\(A=\left(1+5\right)+5^2\left(5+1\right)+5^4\left(5+1\right)+...+5^{28}\left(5+1\right)\)
\(A=6+5^2.6+5^4.6+...+5^{28}.6\)
Vậy A chia hết cho 6
b) (1+3+3^2+3^3+...+3^29) chia hết cho 13
Đặt B= (1+3+3^2+3^3+...+3^29)
\(B=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+...+\left(3^{27}+3^{28}+3^{29}\right)\)
\(B=13+3^3\left(1+3+3^2\right)+....+3^{27}\left(1+3+3^2\right)\)
\(B=13+3^3.13+....+3^{27}.13\)
Vậy B chia hết 13
Câu c,d tương tự.Chúc bạn học tốt
27 tháng 12 2017
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28 tháng 12 2017
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TT
1 tháng 10 2017
Bài 1 : \(A=1+3+3^2+...+3^{31}\)
a. \(A=\left(1+3+3^2\right)+...+3^9.\left(1.3.3^2\right)\)
\(\Rightarrow A=13+3^9.13\)
\(\Rightarrow A=13.\left(1+...+3^9\right)\)
\(\Rightarrow A⋮13\)
b. \(A=\left(1+3+3^2+3^3\right)+...+3^8.\left(1+3+3^2+3^3\right)\)
\(\Rightarrow A=40+...+3^8.40\)
\(\Rightarrow A=40.\left(1+...+3^8\right)\)
\(\Rightarrow A⋮40\)
1 tháng 10 2017
Bài 2:
Ta có: \(C=3+3^2+3^4+...+3^{100}\)
\(\Rightarrow C=(3+3^2+3^3+3^4)+...+(3^{97}+3^{98}+3^{99}+3^{100})\)
\(\Rightarrow3.(1+3+3^2+3^3)+...+3^{97}.(1+3+3^2+3^3)\)
\(\Rightarrow3.40+...+3^{97}.40\)
Vì tất cả các số hạng của biểu thức C đều chia hết cho 40
\(\Rightarrow C⋮40\)
Vậy \(C⋮40\)
9 tháng 9 2017
*Ta có: A\(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=\left(2+2^2\right)+2^2\times\left(2+2^2\right)+...+2^{2008}\times\left(2+2^2\right)\)
\(=\left(2+2^2\right)\times\left(1+2^2+2^3+...+2^{2008}\right)\)
\(=6\times\left(2^2+2^3+...+2^{2008}\right)\)
\(=3\times2\times\left(2^2+2^3+...+2^{2008}\right)\)
\(\Rightarrow A⋮3\)
*Ta có: A \(=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(=2\times\left(1+2+2^2\right)+2^4\times\left(1+2+2^2\right)+...+2^{2008}\times\left(1+2+2^2\right)\)
\(=\left(1+2+2^2\right)\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(=7\times\left(2+2^4+2^7+...+2^{2008}\right)\)
\(\Rightarrow A⋮7\)
Mình sửa lại đề C 1 chút xíu
*Ta có: C \(=3^1+3^2+3^3+3^4+...+3^{2010}\)
\(=\left(3+3^2\right)+3^2\times\left(3+3^2\right)+...+3^{2008}\times\left(3+3^2\right)\)
\(=\left(3+3^2\right)\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=12\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(=4\times3\times\left(1+3^2+3^3+...+3^{2008}\right)\)
\(\Rightarrow C⋮4\)
Các câu khác làm tương tự nhé. Chúc bạn học tốt!
ta có: A = 1 + 3 + 32 +...+ 329 ( có 30 chữ số)
A = (1+3+32) + ...+ (327+328+329) ( có 10 cặp)
A = 13 + ...+ 327.(1+3+32)
A = 13.(1+...+ 327) chia hết cho 13