Induction Proof: Am I on the right track?

[elizaburlap]

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!

 

[tiny-tim]

welcome to pf!

hi elizaburlap! welcome to pf!

(try using the X² and X₂ buttons just above the Reply box )

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)) …

yes, that's all (difficult to read , but) fine!

 

[elizaburlap]

Thank you! This was my first attempt at an induction proof, so I wasn't too sure.

Oh! I see the x₂ now, thanks :)

 

[MrZootSuit]

Math1115. (:p)