Question

Rezolvarea exercitiului: (4 la puterea n+1 × 3 la puterea 2n+2 + 2 la puterea 2n × 9 la puterea n+2 ) ÷ 6 la puterea 2n. Rezolvare in timpul vacantei de iarna, repede!!!!!!

Answer 1

[4^(n+1)*3^(2n+2)+2^(2n)*9^(n+2)]/6^2n=117

[4^(n+1)*3^(2n+2)+2^(2n)*9^(n+2)]/6^2n=
 = [2^(2n+2)*3^(2n+2)+2^(2n)*9^(n+2)]/6^2n=
 =[6^(2n+2)+2^(2n)*9^(n+2)]/6^2n=
 =[(6^2)^(n+1)+2^(2n)*9^(n+2)]/6^2n=
= [36^(n+1)+2^(2n)*9^(n+2)]/6^2n= 
=36^(n+1)/36^n+2^(2n)*9^(n+2)]/6^2n=36+[2^2n*3^(2n+4)/(2^2n*3^2n)]=
=36+1*3^(2n+4-2n)=36+3^4=36+81=117