Chứng minh: 7^1+7^2+7^3+...+7^117+7^118 chia hết cho 57
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=2+2^2+2^3+...+2^{60}\)
\(=2\left(1+2+2^2\right)+...+2^{58}\left(1+2+2^2\right)\)
\(=7\cdot\left(2+...+2^{58}\right)⋮7\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(D=7^1+7^2+7^3+7^4+...+7^{2010}\\ D=\left(7^1+7^2\right)+\left(7^3+7^4\right)+...+\left(7^{2009}+7^{2010}\right)\\ D=7\left(1+7\right)+7^3\left(1+7\right)+...+7^{2009}\left(1+7\right)\\ D=8\left(7+7^3+...+7^{2009}\right)⋮8\\ =>D⋮8->\left(a\right)\\ D=7^1+7^2+7^3+7^4+...+7^{2010}\\ D=\left(7^1+7^2+7^3\right)+\left(7^4+7^5+7^6\right)+...+\left(7^{2008}+7^{2009}+7^{2010}\right)\\ D=7\left(1+7+49\right)+7^4\left(1+7+49\right)+...+7^{2008}\left(1+7+49\right)\\ D=57\left(7+7^4+...+7^{2008}\right)⋮57\\ =>D⋮57->\left(b\right)\\ Từ\left(a\right),\left(b\right)=>D⋮8;D⋮57\)
![](https://rs.olm.vn/images/avt/0.png?1311)
A=7+72+73+...+72016
=(7+72)+(73+74)+...+(72015+72016)
=7.(1+7)+73.(1+8)+...+72015.(1+7)
=7.8+73.8+...+72015.8
=8.(7+73+...+72015) chia hết cho 8 (đpcm)
A=7+72+73+...+72016
=(7+72+73)+...+(72014+72015+72016)
=7.(1+7+72)+...+72014.(1+7+72)
=7.57+...+72014.57
=57.(7+...+72014) chia hết cho 57 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
+) C=5+52+53+54+....+52010
<=> C=(5+52)+(53+54)+.....+(52009+52010)
<=> C=5(1+5)+53(1+5)+....+52009(1+5)
<=> C=5 x 6 +53 x 6+....+52009 x 6
<=> C=6(5+53+....+52009)
=> C chia hết cho 6 (đpcm)
+) C=5+52+53+54+....+52010
<=> C=(5+52+53)+(54+55+56)+....+(52008+52009+52010)
<=> C=5(1+5+25)+54(1+5+25)+....+52008(1+5+25)
<=> C=5 x 31+54x31 +....+52008 x 31
<=> C=31(5+54+....+52008)
=> C chia hết cho 31 (đpcm)
+) D=7+72+73+74+....+72010
<=> D=(7+72)+(73+74)+....+(72009+72010)
<=> D=7(1+7)+73(1+7)+....+72009(1+7)
<=> D=7 x 8 +73 x 8 +....+72009 x 8
<=> D=8(7+73+....+72009)
+) D=7+72+73+74+....+72010
<=> D=(7+72+73)+(74+75+76)+....+(72008+72009+72010)
<=> D=7(1+7+49)+74(1+7+49)+....+72008(1+7+49)
<=> D=7 x 57 +74 x 57+....+72008 x 57
<=> D=57(7+74+...+72008)
=> D chia hết cho 57 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=7\left(1+7+7^2\right)+...+7^{88}\left(1+7+7^2\right)\)
\(=57\left(7+...+7^{88}\right)⋮57\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A=7\left(1+7+7^2\right)+7^4\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)
\(=57\left(7+7^4+...+7^{118}\right)⋮57\)
\(A=7\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)
\(=57\left(7+...+7^{118}\right)⋮57\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=7\left(1+7+7^2\right)+...+7^{118}\left(1+7+7^2\right)\)
\(=57\left(7+...+7^{118}\right)⋮57\)
![](https://rs.olm.vn/images/avt/0.png?1311)
*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!
\(=7\left(1+7+7^2\right)+...+7^{115}\left(1+7+7^2\right)+118\)
\(=57\left(7+...+7^{115}\right)+7^{118}⋮57\)