Calculus Recurence

Question

Find the solution to the recurrence relation an =  4an−1+5an−24an-1+5an-2  with initial terms a0= 2 and a1= 4

Answer 1

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To find the solution to the given recurrence relation \(a_n = 4a_{n-1} + 5a_{n-2} - 4a_{n-1} + 5a_{n-2}\) with initial terms \(a_0 = 2\) and \(a_1 = 4\), we can start by simplifying the recurrence relation:

\[a_n = 4a_{n-1} + 5a_{n-2} - 4a_{n-1} + 5a_{n-2} = 0a_{n-1} + 10a_{n-2} = 10a_{n-2}.\]

So, we have simplified the recurrence relation to \(a_n = 10a_{n-2}\).

Now, let's find a few terms of the sequence to see the pattern:

\(a_0 = 2\)

\(a_1 = 4\)

\(a_2 = 10a_0 = 10 \cdot 2 = 20\)

\(a_3 = 10a_1 = 10 \cdot 4 = 40\)

\(a_4 = 10a_2 = 10 \cdot 20 = 200\)

\(a_5 = 10a_3 = 10 \cdot 40 = 400\)

\(a_6 = 10a_4 = 10 \cdot 200 = 2000\)

It appears that \(a_n = 10^n \cdot 2\) for \(n \geq 0\).

So, the solution to the recurrence relation with initial terms \(a_0 = 2\) and \(a_1 = 4\) is:

\[a_n = 10^n \cdot 2\]