## C=4+4^2+4^3+4^4+……+4^2014+4^2015+4^2016 Chứng minh c chia hết cho 21 và 105

Question

C=4+4^2+4^3+4^4+……+4^2014+4^2015+4^2016
Chứng minh c chia hết cho 21 và 105

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3 tháng 2021-09-15T03:33:47+00:00 2 Answers 7 views 0

$C=4+4^2+4^3+4^4+……+4^{2014}+4^{2015}+4^{2016}\\=(4+4^2+4^3)+……+(4^{2014}+4^{2015}+4^{2016})\\=4.(1+4+4^2)+…+4^{2014}.(1+4+4^2)\\=4.21+…+4^{2014}.21\\=21.(4^2+…+4^{2014}) \vdots 21\\$___________________________$\\C=4+4^2+4^3+4^4+……+4^{2014}+4^{2015}+4^{2016}\\=(4+4^2+4^3+4^4+4^5+4^6)+……+(4^{2011}+4^{2012}+4^{2013}4^{2014}+4^{2015}+4^{2016})\\=4.(1+4+4^2+4^3+4^4+4^5)+…+4^{2011}.(1+4+4^2+4^3+4^4+4^5)\\=1365.4+…+1365.4^{2011}\\=1365.(4+..+4^{2011})\vdots 105$
C =4 + $4^{2}$+ $4^{3}$+ $4^{4}$+ ….+$4^{2014}$+ $4^{2015}$+ $4^{2016}$
=( 4 + $4^{2}$+ $4^{3}$)+($4^{4}$+$4^{5}$+$4^{6}$ )$+ ….+($4^{2014}$+$4^{2015}$+$4^{2016}$) =4 .( 1 + 4 +$4^{2}$) +$4^{4}$.( 1 + 4 +$4^{2}$) + ….. +$4^{2014}$.( 1 + 4 +$4^{2}$) =4 . 21 +$4^{4}$. 21 +…..+$4^{2014}$. 21 =21.(4 +$4^{4}$+……+$4^{2014}$) chia hết cho 21 _____________________________________________________________________________ C =4 +$4^{2}$+$4^{3}$+$4^{4}$+ ….+$4^{2014}$+$4^{2015}$+$4^{2016}$=( 4 +$4^{2}$)+ ($4^{3}$+$4^{4}$)+ ….+($4^{2015}$+$4^{2016}$) =4 .( 1 + 4 ) +$4^{3}$.( 1 + 4 ) + ….. +$4^{2015}$.( 1 + 4 ) =4 . 5 +$4^{4}$. 5 +…..+$4^{2014}$. 5 =5.(4 +$4^{3}$+……+$4^{2015}$) chia hết cho 5 Vì C=4 +$4^{2}$+$4^{3}$+$4^{4}$+ ….+$4^{2014}$+$4^{2015}$+$4^{2016}\$ chia hết cho 21;5 nên C chia hết cho 105