Question

What is the recursive for of the following formula of a sequence? an=8(2)^n-1?a) a1=2, an=8an-1(n-1)b) a1=8, an=2an-1+8c) a1=2, an=8an-1d) a1=8, an=2an-1

Answer 1

Answer:

d) a1 = 8, an = 2a(n-1)

Explanation:

Let's see the first 3 terms of the sequence

$\begin{gathered} a_n=8(2)^{n-1} \\ a_1=8(2)^{1-1}=8(2)^0=8 \\ a_2=8(2)^{2-1}=8(2)^1=8(2)=16 \\ a_3=8(2)^{3-1}=8(2)^2=8(4)=32 \end{gathered}$

Therefore, the first term is a1 = 8

And the next terms are twice the term before, we get

a2 = 2(8) = 16

a3 = 2(16) = 32

Then, the recursive formula is equal to

d) a1 = 8, an = 2a(n-1)