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Tarasdaniela010
a fost răspuns

Aratati ca nr A=7*12^n3^n+1+6*4^n+1*9^n+2+18^n+1*2^n+1 este divizibil cu 2001, ptr orice nr nat n

Răspuns :

Renatemambouko
A=7*2^12n*3^n+1+6*4^n+1*9^n+2+18^n+1*2^n+1 =
=7*2^2n*3^n*3^n+1+6*2^2n+2*3^2n+4+2^n*3^2n+1*2^n+1 =
=7*2^2n*3^2n+1+6*2^2n+2*3^2n+4+2^2n+2*3^2n+2 =
=2^2n*3^2n(7*3+6*2^2*3^4+2^2*3^2)=
=2^2n*3^2n(21+6*4*81+4*9)=
=2^2n*3^2n(21+1944+36)=
=2^2n*3^2n * 2001  este divizibil cu 2001

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