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Ta có: 31+32+33+34+35+...+32012
=(3^1+3^2+3^3+3^4+3^5)+...+(3^2008+3^2009+^3^2010+3^2011+3^2012)
=(3*1+3*3+3*3^2+3*3^3+3*3^4)+...+(3^2008*1+3^2008*3+3^2008*3^2+3^2008*3^3+3^2008*3^4)
=3*(1+3+3^2+3^3+3^4)+....+3^2008*(1+3+3^2+3^3+3^4)
=3*121+...+3^2008*121
=(3+3^6+...+3^2008)*121
Vì 121 chia 120 dư 1
Nên 31+32+33+34+35+...+32012 chia hết cho 120
*là nhân nha bạn
Đặt S=\(3\)\(+\)\(3^2\)\(+\)\(3^3\)\(+\)...............\(+\)\(3^{2012}\)
\(\Rightarrow\)S=[\(3\)\(+\)\(3^2\)\(+\)\(3^3\)\(+\)]\(+\)........................\(+\)[\(3^{2009}\)\(+\)\(3^{2010}\)\(+\)\(3^{2011}\)\(+\)\(3^{2012}\)]
\(\Rightarrow\)S=120\(+\).......................\(+\)\(3^{2008}\)[\(3\)\(+\)\(3^2\)\(+\)\(3^3\)\(+\)\(3^4\)]
\(\Rightarrow\)S=120\(+\).......................\(+\)\(3^{2008}\)\(+\)120
\(\Rightarrow\)S=120[1\(+\)................\(+\)\(3^{2008}\)]
VÌ 120\(⋮\)120 \(\Rightarrow\)S\(⋮\)120
ta có 3^1+3^2+........+3^2012
=>(3^1+3^2+3^3+3^4)+.........+3^2009(3^1+3^2+3^3+3^4)
=>120+........................................+3^2009*120
=>120*(1+...............+3^2009) chia hết cho 120
vậy 3^1+3^2+.............+3^1012 chia hết cho 120
\(3^1+3^2+3^3+3^4+...+3^{2012}\)
\(=\left(3^1+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(=\left(3^1+3^2+3^3+3^4\right)+3^4\left(3^1+3^2+3^3+3^4\right)+...+3^{2008}\left(3^1+3^2+3^3+3^4\right)\)
\(=120\left(1+3^4+...+3^{2008}\right)\)chia hết cho \(120\).
\(3+3^2+3^3+...+3^{2012}\)
\(=\left(3+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)\)
\(=\left(3+3^2+3^3+3^4\right)+3.\left(3+3^2+3^3+3^4\right)+...+3^{2008}\left(3+3^2+3^3+3^4\right)\)
mà \(3+3^2+3^3+3^4=3+9+27+81=120⋮120\)
\(\Rightarrow\hept{\begin{cases}3+3^2+3^3+3^4⋮120\\3\left(3+3^2+3^3+3^4\right)⋮120\\3^{2008}\left(3+3^2+3^3+3^4\right)⋮120\end{cases}.......}\)
\(\Rightarrow3+3^2+3^3+...+3^{2012}⋮120\)
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6-9369--959295-2===
a) ta có A= 2+2^2+2^3+2^4+2^5+2^6
=2*(1+2+2^2+2^3+2^4+2^5)
=2*63 =2*21*3 CHIA HẾT CHO 3( vì có một thứa số 3 trong tích )
còn lại bạn làm tương tự nha
Ta có : \(3^1=3;3^2=9;3^3=27;3^4=81\)
Do đó : \(3^1+3^2+3^3+3^4=3+9+27+81=120\)
Nên \(3^1+3^2+3^3+3^4+3^5+...+3^{2012}=\left(3^1+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)+...+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)=120+3^4.120+...+3^{2008}.120=120.\left(1+3^4+...+3^{2008}\right)⋮120\) Vậy tổng \(S=3^1+3^2+3^3+...+3^{2012}⋮120\)
Ta có :
\(3^1=3;3^2=9;3^3=27;3^4=81;3^5=243\)
Do đó :
\(3^1+3^2+3^3+3^4+3^5=3+9+27+81+243=363\)
Nên
\(3^1+3^2+3^3+3^4+3^5+....+3^{2012}=\left(3^1+3^2+3^3+3^4\right)+\left(3^5+3^6+3^7+3^8\right)\)\(+.......+\left(3^{2009}+3^{2010}+3^{2011}+3^{2012}\right)=120+3^4.120+......3^{2008}.120\)
Vậy \(3^1+3^2+3^3+3^4+3^5+.....+3^{2012}\)không chia hết cho 120
Tổng trên chia hết cho 120 vì
\(\left(3+3^2+3^3+3^4\right)=120\)
thế nên cứ tổng 4 số hạng liên tiếp của tổng trên là chia hết cho 120
mà 120 chia hết cho 4
nên tổng đã cho chia hết cho 120