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a, Ta có: A = 3 + 3^2 + 3^3 + ... + 3^99 + 3^100
=> 3A = 3( 3 + 3^2 + 3^3 + ... + 3^99 + 3^100)
=> 3A = 3. 3 + 3. 3^2 + 3. 3^3 + ... + 3. 3^99 + 3. 3^100
=> 3A = 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^101
=> 3A - A = ( 3^2 + 3^3 + 3^4 + ... + 3^100 + 3^101 ) - ( 3 + 3^2 + 3^3 + ... + 3^99 + 3^100 )
=> 2A = 3^101 - 3
=> A = \(\dfrac{3^{101}-3}{2}\)
Vậy dạng viết gọn của A là: \(\dfrac{3^{101}-3}{2}\)
b, Ta có: A = 3 + 3^2 + 3^3 + ... + 3^99 + 3^100
=> A = ( 3 + 3^2 ) + ( 3^3 + 3^4 ) + ... + ( 3^99 + 3^100 )
=> A = 3( 1 + 3 ) + 3^3 ( 1 + 3 ) + ... + 3^99( 1 + 3 )
=> A = 3. 4 + 3^3. 4 + ... + 3^99. 4
=> A = 4( 3 + 3^3 + ... + 3^99 ) chia hết cho 4
=> A chia hết cho 4
Vậy A chia hết cho 4 ( điều phải chứng minh )
Chúc bạn hoc tốt! ~
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*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!
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2^1+2^2+2^3+2^4+...+2^{2010}\)
\(A=\left(2^1+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(A=7\left(2^1+2^4+...+2^{2008}\right)⋮7\)
Các ý dưới bạn làm tương tự nhé.
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 1:
A=400x7x36+1620
*400x7x36 \(⋮\)2;3;5;9
1620 \(⋮\) 2;3;5;9
\(\Rightarrow\)400x7x36+1620\(⋮\) 2;3;5;9
Bài 2:
C=3+32+33+........+360
=(3+32)+(33+34)+...........+(359+360)
=3.(1+2)
Bài 2 :
a, \(C=3+3^2+3^3...+3^{60}\)
\(\Rightarrow C=\left(3+3^2\right)+\left(3^3+3^4\right)+...\left(3^{59}+3^{60}\right)\)
\(\Rightarrow C=1\left(1+3\right)+3^3\left(1+3\right)+..+3^{59}\left(1+3\right)\)
\(\Rightarrow C=4.\left(1+3^3+...+3^{59}\right)\)
\(\Rightarrow C⋮4\)
\(b,1+3+3^2+3^3+...+3^{60}\)
\(\Rightarrow3A=3+3^2+3^3+...+3^{60}+3^{61}\)
\(\Rightarrow3A-A=\left(3+3^2+3^3..+3^{60}+3^{61}\right)-\left(1+3+3^2+...+3^{60}\right)\)
\(\Rightarrow2A=3^{61}-1\)
\(\Rightarrow A=\frac{3^{61}-1}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt biểu thức trên là A
Chứng minh A\(⋮4\)
Ta có :A=\(3+3^2+3^3+...+3^{59}+3^{60}\)
A=\(\left(3+3^2\right)+\left(3^3+3^4\right)+...+\left(3^{59}+3^{60}\right)\)
A=\(3\left(1+3\right)+3^3\left(1+3\right)+...+3^{59}\left(1+3\right)\)
A=\(3.4+3^3.4+...+3^{59}.4\)
A=\(4\left(3+3^3+...+3^{59}\right)\)
Vậy \(A⋮4\)
Chứng minh \(A⋮13\)
Ta có :A=\(3+3^2+3^3+...+3^{59}+3^{60}\)
A=\(\left(3+3^2+3^3\right)+...+\left(3^{58}+3^{59}+3^{60}\right)\)
A=\(3\left(1+3+3^2\right)+...+3^{58}\left(1+3+3^2\right)\)
A=\(3.13+...+3^{58}.13\)
A=\(13\left(3+...+3^{58}\right)\)
Vậy \(A⋮13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Mình ngu lắm dân trần đăng ninh chuyên anh mà làm sao giỏi toán được
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=\left(1+3+3^2\right)+...+\left(3^{99}+3^{100}+3^{101}\right)\\ A=\left(1+3+3^2\right)+...+3^{99}\left(1+3+3^2\right)\\ A=\left(1+3+3^2\right)\left(1+...+3^{99}\right)=13\left(1+...+3^{99}\right)⋮13\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=(1+2010)+2010 mũ 2+2010 mũ 3 +...+2010 mũ 6 + 2010 mũ 7
A=2011+2010 mũ 2(1+2010)+...+2010 mũ 6(1+2010)
A=2011+2010 mũ 2.2011+...2010 mũ 6.2011
A=2011(1+2010+...+2010 mũ 6)chia hết cho 2011
A = 3+33+....+32010
đề bài đó hả
Nếu vậy thi \(3+3^3+3^5+....+3^{2009}\) chứ, 3^2010 là sao mà hợp sãy số