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

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

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