CHứng minh :
A=1+3+3^2+3^3+……+3^11 chia hết cho 13 , chia hết cho 40 .
B = 4 +4^2+4^3+4^4+……….+4^23+4^24 chia hết cho 20, chia hết cho 21.
CHứng minh :
A=1+3+3^2+3^3+……+3^11 chia hết cho 13 , chia hết cho 40 .
B = 4 +4^2+4^3+4^4+……….+4^23+4^24 chia hết cho 20, chia hết cho 21.
Đáp án:
Giải thích các bước giải:
Ta có: \(A=1+3+3^2+…+3^{11}\)
\(=\left(1+3+3^2\right)+\left(3^3+3^4+3^5\right)+\left(3^6+3^7+3^8\right)+\left(3^9+3^{10}+3^{11}\right)\)
\(=\left(1+3+3^2\right)+3^3\left(1+3+3^2\right)+3^6\left(1+3+3^2\right)+3^9\left(1+3+3^2\right)\)
\(=13+3^3.13+3^6.13+3^9.13\)
\(=13\left(1+3^3+3^6+3^9\right)⋮13\)
\(\Rightarrow A⋮13\)
Ta lại có:\(A=1+3+3^2+…+3^{11}\)
\(=\left(1+3+3^2+3^3\right)+\left(3^4+3^5+3^6+3^7\right)+\left(3^8+3^9+3^{10}+3^{11}\right)\)
\(=\left(1+3+3^2+3^3\right)+3^4\left(1+3+3^2+3^3\right)+3^8\left(1+3+3^2+3^3\right)\)
\(=40+3^4.40+3^8.40\)
\(=40\left(1+3^4+3^8\right)⋮40\)
\(\Rightarrow A⋮40\)
Ta có :\(B=4+4^2+…..+4^{23}+4^{24}\)
\(=\left(4+4^2\right)+\left(4^3+4^4\right)+….+\left(4^{23}+4^{24}\right)\) (12 nhóm)
\(=4\left(4+4^2\right)+4^3\left(4+4^2\right)+…….+4^{23}\left(4+4^2\right)\)
\(=4.20+4^3.20+…..+4^{23}.20\)
\(=20\left(4+4^3+…+4^{23}\right)⋮20\)
\(\Rightarrow B⋮20\)
Ta có :\(B=4+4^2+4^3+……..+4^{23}+4^{24}\)
\(=\left(4+4^2+4^3\right)+\left(4^4+4^5+4^6\right)+…….+\left(4^{22}+4^{23}+4^{24}\right)\)
\(=4\left(1+4+4^2\right)+4^4\left(1+4+4^2\right)+….+4^{22}\left(1+4+4^2\right)\)
\(=4.21+4^4.21+….+4^{22}.21\)
\(=21\left(4+4^4+……+4^{22}\right)⋮21\)
\(\Rightarrow B⋮21\)
1,$A=1+3+3^2+3^3+…+3^{11}$
$=(1+3+3^2)+(3^3+3^4+3^5)+….+(3^9+3^{10}+3^{11})$
$=13+3^3.(1+3+3^2)+…+3^9(1+3+3^2)$
$=13+13.3^3+…+13.3^9$
$=13(1+3^3+…+3^9) \vdots 13$
$A=1+3+3^2+3^3…+3^{11}$
$=(1+3+3^2+3^3)+(3^4+3^5+3^6+3^7)+(3^8+3^9+3^{10}+3^{11})$
$=40+3^4(1+3+3^2+3^3)+3^8(1+3+3^2+3^3)$
$=40+3^4.40+3^8.40$
$=40.(1+3^4+3^8) \vdots 40$
$B=4+4^2+4^3…+4^{24}$
$=(4+4^2)+(4^3+4^4)+…+(4^{23}+4^{24})$
$=4.(1+4)+4^3.(1+4)+…+4^{23}.(1+4)$
$=4.5+4^3.5+…+4^{23}.5$
$=5.(4+4^3+…+4^{23}) \vdots 5$
Mà $B=4+4^2+4^3…+4^{24} \vdots 4$
$⇒B \vdots 4.5=20$ (do $(4;5)=1$)
$B=4+4^2+4^3…+4^{24}$
$=(4+4^2+4^3)+(4^4+4^5+4^6)+…+(4^{22}+4^{23}+4^{24})$
$=4.(1+4+4^2)+4^4(1+4+4^2)+…+4^{22}.(1+4+4^2)$
$=4.21+4^4.21+…+4^{22}.21$
$=21.(4+4^4+…+4^{22}) \vdots 21$