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

0 bình luận về “C=4+4^2+4^3+4^4+……+4^2014+4^2015+4^2016 Chứng minh c chia hết cho 21 và 105”

1. Đáp án+Giải thích các bước giải:

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

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2.           #Dqnghiep

   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

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   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

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