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... tìm số dư khi chia hết???
nếu nó chia hết thì số dư bằng 0 rồi
Trời trời, mình làm cho bạn câu khi nãy bạn phải biết vận dụng cho mấy bài sau chứ, câu này giống i lột câu khi nãy luôn ấy, nhưng thôi, khá rảnh nên:vv
+Ta có: \(B=3+3^2+3^3+3^4+...+3^{2010}\)
-> \(B=3\left(1+3\right)+3^3\left(1+3\right)+...+3^{2009}\left(1+3\right)\)
-> \(B=3.4+3^3.4+...+3^{2009}.4\)
-> \(B=4\left(3+3^3+...+3^{2009}\right)⋮4\)
-> Đpcm
+ Ta có: \(B=3+3^2+3^3+3^4+....+3^{2010}\)
-> \(B=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+...+3^{2008}\left(1+3+3^2\right)\)
-> \(B=3.13+3^4.13+...+.3^{2008}.13\)
-> \(B=13\left(3+3^4+...+3^{2008}\right)⋮13\)
-> Đpcm
Ta có: \(B=3^1+3^2+3^3+3^4+...+3^{2010}\)
\(=3^1\cdot\left(1+3\right)+3^3\cdot\left(1+3\right)+...+3^{2009}\cdot\left(1+3\right)\)
\(=\left(1+3\right)\cdot\left(3^1+3^3+...+3^{2009}\right)\)
\(=4\cdot\left(3+3^3+...+3^{2009}\right)⋮4\)(đpcm)
Ta có: \(B=3^1+3^2+3^3+3^4+...+3^{2010}\)
\(=3\left(1+3+3^2\right)+3^4\cdot\left(1+3+3^2\right)+...+3^{2008}\cdot\left(1+3+3^2\right)\)
\(=\left(1+3+3^2\right)\cdot\left(3+3^4+...+3^{2008}\right)\)
\(=13\cdot\left(3+3^4+...+3^{2008}\right)⋮13\)(đpcm)
Úi gời cơi cộng chấm chấm chấm :)))
+ Ta có: \(A=2+2^2+2^3+2^4+...+2^{2010}\)
\(A=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(A=2.3+2^3.3+...+2^{2009}.3\)
\(A=3\left(2+2^3+...+2^{2010}\right)⋮3\)
-> Đpcm
+ Ta có: \(A=2+2^2+2^3+2^4+...+2^{2010}\)
\(A=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+....+2^{2008}\left(1+2+2^2\right)\)
\(A=2.7+2^4.7+...+2^{2008}.7\)
\(A=7\left(2+2^4+...+2^{2008}\right)⋮7\)
-> Đpcm
\(A=2^1+2^2+...+2^{2010}\)
\(=\left(2^1+2^2\right)+\left(2^3+2^4\right)+...+\left(2^{2009}+2^{2010}\right)\)
\(=2\left(1+2\right)+2^3\left(1+2\right)+...+2^{2009}\left(1+2\right)\)
\(=3\left(2+2^3+...+2^{2009}\right)⋮3\)
\(A=2+2^2+2^3+...+2^{2010}\)
\(=\left(2+2^2+2^3\right)+\left(2^4+2^5+2^6\right)+...+\left(2^{2008}+2^{2009}+2^{2010}\right)\)
\(=2\left(1+2+2^2\right)+2^4\left(1+2+2^2\right)+...+2^{2008}\left(1+2+2^2\right)\)
\(=7\left(2+2^4+...+2^{2008}\right)⋮7\)
A=2\(^1\)+2\(^2\)+...+2\(^{2010}\)
=(2\(^1\)+2\(^2\))+(2\(^3\)+2\(^4\))+...+(2\(^{2009}\)+2\(^{2010}\))
=2(1+2)+2\(^3\)(1+2)+...+2\(^{2009}\)(1+2)
=3(2+2\(^3\)+...+2\(^{2009}\))⋮3