Question

Solve this recurrence relation together with the initial condition given. an = 2an−1 for n ≥ 1, a0 = 3

Answer 1

The solution of the recurrence relation is $a_n=3.2^n$

For given question,

We have been given a recurrence relation $a_n = 2a_{n-1}$ for n ≥ 1

and an initial condition $a_0=3$

Let $a_n$ = m², $a_{n-1}$ = m and $a_{n-2}$ = 1

So from given recurrence relation we get an characteristic equation,

⇒ m² = 2m

⇒ m² - 2m = 0                     .........( Subtract 2m from each side)

⇒ m(m - 2) = 0                     .........(Factorize)

⇒ m = 0    or  m - 2 = 0

⇒ m = 0   or   m = 2

We know that the solution of the recurrence relation is then of the form

$a_n=\alpha_1 {m_1}^n + \alpha_2 {m_2}^n$  where $m_1,m_2$ are the roots of the characteristic equation.

Let, $m_1$ = 0   and $m_2$ = 2

From above roots,

$\Rightarrow a_n=\alpha_1 {0}^n + \alpha_2 {2}^n\\\\\Rightarrow a_n=0+\alpha_2 {2}^n\\\\\Rightarrow a_n=\alpha_2 {2}^n$

For n = 0,

$\Rightarrow a_0=\alpha_2 {2}^0\\\\\Rightarrow a_0=\alpha_2 \times 1\\\\\Rightarrow a_0=\alpha_2$

But  $a_0=3$

This means $\alpha_2=3$

so, the solution of the recurrence relation would be $a_n=3.2^n$

Therefore, the solution of the recurrence relation is $a_n=3.2^n$

Learn more about the recurrence relation here:

brainly.com/question/27618667

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