Answer to Question #270690 in Discrete Mathematics for nur

Question

Consider a recurrence relation a_n = 3a_n-1 for n = 1,2,3,4,... and suppose a₁ =4. Calculate a₅.

Answer 1

$a_5=3a_4=3(3a_3)=9(3a_2)=27(3a_1)=81(4)=324$

We have geometric sequence

$a=a_1=4, q=\dfrac{a_n}{a_{n-1}}=3, n\geq2$

Then

$a_n=a(q)^{n-1}$

$a_5=4(3)^{5-1}=324$