a,Ta có: A có 2016 số số hạng, ghép A thành 504 nhóm, mỗi nhóm có 4 số hạng như sau :
\(A=(3+3^2+3^3+3^4)+(3^5+3^6+3^7+3^8)+....+(3^{2013}+3^{2014}+3^{2015}+3^{2016})\)
\(A=3.(1+3+3^2)+3^5.(1+3+3^2)+....+3^{2013}.(1+3+3^2)\)
\(A=3.13+3^5.13+....+3^{2013}.13\)
\(A=13.(3+3^5+...+3^{2013})⋮13\)
\(\Rightarrow A⋮13\)
\(a\)) Ta có :
\(A=3+3^2+3^3+..........+3^{2016}\) (2016 số hạng )
\(A=\left(3+3^2+3^3\right)+\left(3^4+3^5+3^6\right)+.....+\left(3^{2014}+3^{2015}+3^{2016}\right)\) (672 nhóm )
\(A=3\left(1+3+3^2\right)+3^4\left(1+3+3^2\right)+.......+3^{2015}\left(1+3+3^2\right)\)
\(A=3.13+3^4.13+........+3^{2015}.13\)
\(A=13\left(3+3^4+.......+3^{2016}\right)\)
\(\Rightarrow A\) \(⋮\) \(13\)
\(\Rightarrowđpcm\)
\(b\)) Ta có :
\(A=3+3^2+3^3+..........+3^{2016}\)
\(\Rightarrow3A=3^2+3^3+...............+2^{2016}+3^{2017}\)
\(\Rightarrow3A-A=3^{2017}-3\)
\(\Rightarrow2A=3^{2017}-3\)
\(\Rightarrow2A+3=3^{2017}\)(1)
Theo bài ta có :
\(2A+3=3^{2x}\)(2)
Từ (1) và (2) ta có :
\(3^{2x}=3^{2017}\)
\(\Rightarrow2x=2017\)
\(x=2017:2\)
\(x=1008,5\) ( ko thoả mãn \(x\in N\))
Vậy ko tìm dc giá trị của \(x\) thỏa mãn theo yêu cầu
a) A=3+32+33+...+32016
A=(3+32+33)+(34+35+36)+...+(32014+32015+32016)
A=3.(1+3+32)+34.(1+3+32)+...+32014.(1+3+32)
A=3.13+34.13+...+32014.13
A=13.(3+34+...+32014)
Vì A có chứa thừa số 13 nên A chia hết cho 13
b) A=3+32+33+...+32016
3A=3.(3+32+33+...+32016)
3A=32+33+...+32017
3A-A=(32+33+...+32017)-(3+32+33+...+32016)
2A=32017-3
A=\(\dfrac{3^{2017}-3}{2}\)
2A+3=32x
2.\(\dfrac{3^{2017}-3}{2}\)+3=32x
32017-3=32x
32017=32x:3=32x-1
=>2x-1=2017
2x=2017+1=2018
x=2018:2=1009
b, \(A=3+3^2+3^3+....+3^{2016}\)
\(3A=3(3+3^2+3^3+...+3^{2016})\)
\(3A=3^2+3^3+3^4+...+3^{2016}+3^{2017}\)
\(2A=3A-A=(3^2+3^3+...+3^{2016}+3^{2017})-(3+3^2+3^3+...+3^{2016})\)\(2A=3^{2017}-3^{ }\)
\(\Rightarrow2A+3=3^{2017}-3+3\)
\(\Rightarrow2A+3=3^{2017}\)
\(\Rightarrow2x=2017\)
\(\Rightarrow2x=1008,5\)
Mà \(x\in Z\) nên suy ra \(x\in\varnothing\)
