Question

URGENT!!!!!!!!!!!!!!

Answer 1

Avem o relatie de recurenta de ordinul 2, deci vom rezolva ecuatia caracteristica. Pt inceput, rescriem relatia de recurenta astfel:
$a_{k+2} -5a_{k+1}+4a_{k}$

Apoi rezolvam ecuatia caracteristica:
[tex]q^2-5q+4=0 [/tex]
Δ=25-16=9
[tex]q_{1}= \frac{5+ \sqrt{9} }{2} [/tex]
$q_{2}= \frac{5- \sqrt{9} }{2}$

Deci q1=4, q2=1.
Termenul general va fi de forma:
$a_{n}=c_{1}q_{1}^n+c_{2}q_{2}^n$

Deci
$a_{n}=c_{1}4^n+c_{2}$

Determinam c1 si c2 din primii 2 termeni ai sirului:
$\left \{ {{4c_{1}+c_{2}=5} \atop {16c_{1}+c2=17}} \right.$

Daca scadem prima ecuatia din a doua, obtinem:
$12c_{1}=12$
Deci c1=1. 4+c2=5, deci c2=1.

In final
$a_{n}=4^n+1$