- #1

elizaburlap

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## Homework Statement

Let a(1)=a(2)=5 and a(n+1)=a(n)+6a(n-1), n≥2

Use induction to prove that a(n)=(3^n)-(-2)^n for n≥1

## Homework Equations

Not applicable

## The Attempt at a Solution

I have check that a(3) = 5+6·5 = 35 = 3^3-(-2)^3 so it holds for n = 3.

The cases n = 1 and n = 2 are similar and also hold

So I assumed that it holds for n and considered

a(n+1) = a(n)+6a(n-1)

= (3^n-(-2)^n)+6(3^(n-1)-(-2)^(n-1))

The second term [6(3^(n-1)-(-2)^(n-1))] equals [2·3^n+3·(-2)^n].

So,

a(n+1) = (3^n-(-2)^n)+(2·3^n+3·(-2)^n)

= 3^(n+1)-(-2)^(n+1)

Is this a valid induction proof? Am I on the right lines here?

Thanks!