Solution

The correct answer is C, D Only

Explanation:

To find which sequences \( {a_n}\) are solutions to the recurrence relation \( a_n = 8a_{n-1} - 16a_{n-2}\), let's analyze each option:

A. an = 1

For constant sequences, the relation would imply an equality like: [\( 1 = 8 \cdot 1 - 16 \cdot 1\)] [ 1 = 8 - 16 ] [ 1 = -8 ] This is clearly false.

B. an = 2n

Let's substitute an = 2n into the recurrence relation: [\( 2^n = 8 \cdot 2^{n-1} - 16 \cdot 2^{n-2}\) ]

=> [ \(2^n = 8 \cdot 2^{n-1} - 16 \cdot 2^{n-2}\) ]

=> [ \(2^n = 8 \cdot 2^{n-1} - 16 \cdot 2^{n-2} = 8 \cdot \frac{2^n}{2} - 16 \cdot \frac{2^n}{4}\) ]

=> [ \(2^n = 4 \cdot 2^n - 4 \cdot 2^n\) ] [ \(2^n = 0\) ] This is also false.

C. an = 4n

Let's check this one:

=> [ \(4^n = 8 \cdot 4^{n-1} - 16 \cdot 4^{n-2}\) ]

=> [ \(4^n = 8 \cdot 4^{n-1} - 16 \cdot 4^{n-2} = 8 \cdot \frac{4^n}{4} - 16 \cdot \frac{4^n}{16}\) ]

=> [ \(4^n = 2 \cdot 4^n - 4^n\) ]

=> [ \(4^n = 4^n\) ] This is true, so ( \(a_n = 4^n\) ) is a solution.

D. an = n4n

Let's check this one:

=> [ \(a_n = n4^n\) ]

=> [ \(n \cdot 4^n = 8(n-1) \cdot 4^{n-1} - 16(n-2) \cdot 4^{n-2}\)

\(= 8(n-1) \cdot 4^{n-1} - 16(n-2) \cdot 4^{n-2}\)

\(= 8(n-1) \cdot \frac{4^n}{4} - 16(n-2) \cdot \frac{4^n}{16}\)

\(= 8(n-1) \times \frac{4^n}{4} - 16(n-2) \cdot \frac{4^n}{16}\) ]

=> [ \( n \cdot 4^n = 2(n-1)4^n - (n-2)4^n\) ]

=> [ \(n \cdot 4^n = 2(n-1)4^n - (n-2)4^n\) ]

=> [ \(n \cdot 4^n = (2n-2- n + 2)4^n\) ]

=> [ \(n \cdot 4^n = n4^n\) ] This relation holds, so ( \(n4^n\) ) is a solution.

E. an = (-4)n

Let's check: [ \((-4)^n = 8 \cdot (-4)^{n-1} - 16 \cdot (-4)^{n-2}\) ]

[ \((-4)^n = 8 \cdot \left(\frac{(-4)^n}{-4}\right) - 16 \cdot \left(\frac{(-4)^n}{16}\right)\) ]

[ \((-4)^n = -2(-4)^n - (-4)^n\) ]

[ \((-4)^n = -3(-4)^n\) ]

[ \((-4)^n = -3(-4)^n\) ] This is invalid, so it is also a solution.

Given the solutions, the correct option is: C, D Only