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FobelloE

Answered question

2021-09-23

Show that the sequence

a_{n}

is an solution of the recurrence relation

a_{n} = - 3 a_{n - 1} + 4 a_{n - 2} if a_{n} = {(- 4)}^{n}

Nichole Watt

Skilled2021-09-24Added 100 answers

Given:

a_{n} = {(- 4)}^{n}

n=0,1,2,...
To prove:

a_{n} = - 3 a_{n - 1} + 4 a_{n - 2}, n \geq 2

.
Proof
Replace n in

a_{n} = {(- 4)}^{n} b y n - 1 :

a_{n - 1} = {(- 4)}^{n - 1}

Replace n in

a_{n} = {(- 4)}^{n} b y n - 2 :

a_{n - 2} = {(- 4)}^{n - 2}

We will start from the expression

- 3 a_{n - 1} + 4 a_{n - 2}

and prove that this term has to be equal to

a_{n} (w h e n n \geq 2)

.
Let us use the two previous expressions derived for

a_{n - 1} and a_{n - 2}

- 3 a_{n - 1} + 4 a_{n - 2} = - 3 {(- 4)}^{n - 1} + 4 {(- 4)}^{n - 2}

= - 3 \cdot (- 4) \cdot {(- 4)}^{n - 2} + 4 {(- 4)}^{n - 2}

Let us factor out

{(- 4)}^{n - 2} :

= {(- 4)}^{n - 2} (- 3 \cdot (- 4) + 4)

= {(- 4)}^{n - 2} (12 + 4)

= {(- 4)}^{n - 2} (16)

= {(- 4)}^{n - 2} {(- 4)}^{2}

= {(- 4)}^{n}

= a_{n}

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