Question

Rezolvați si mie va rog

Answer 1

la b) la numarator ai 2^(n+3)+2^(n+2)+2^(n+1)+2^n=8*2^n+4*2^n+2*2^n+2^n=(8+4+2+1)*2^n=15*2^n=30*2^(n-1), iar la numitor 3^(n+2)+2*3^(n+1)+3^n=9*3^n+6*3^n+3^n=(9+6+1)*3^n=16*3^n=48*3^(n-1), de unde concluzia.

la c) la numarator ai 4^n*5^(n+1)+4^(n+1)*5^n=5*4^n*5^n+4*4^n*5^n=5*20^n+4*20^n=(4+5)*20^n=9*20^n, iar la numitor 3^n*7^(n+1)+3^(n+1)*7^n-21^n=7*3^n*7^n+3*3^n*7^n-21^n=7*21^n+3*21^n-21^n=(7+3-1)*21^n=9*21^n, de unde concluzia.