prove that recurrence relation is odd

Hello!

I just want to be sure if I did solve the task correctly.

The task is:

Prove that bn is odd for integers n>=1

$\displaystyle b_1=1$
$\displaystyle b_2=3$
$\displaystyle b_k=b_{k-2}+2b_{k-1}$ for k>=3

The attempt at a solution

Induction basis:

$\displaystyle b_1=1$ true 1 is odd
$\displaystyle b_2=3$ true 2 is odd

Now the question is:

Could I use the strong (complete) induction?

If I can use it, the solution is simple:

Let the recurrence relation is true for all k, such that n<k and n=k i.e n<=k
$\displaystyle b_k=b_{k-2}+2b_{k-1}$

then for n=k+1

$\displaystyle b_{k+1}=b_{k-1}+2b_{k}$

$\displaystyle b_{k-1}$ is odd and $\displaystyle 2b_k$ is even therefore odd+even=odd

Is this correct?

Thank you.

Regards.

Quote:

---

Originally Posted by p0oint

Let the recurrence relation is true for all k, such that n<k and n=k i.e n<=k
$\displaystyle b_k=b_{k-2}+2b_{k-1}$

then for n=k+1

$\displaystyle b_{k+1}=b_{k-1}+2b_{k}$

$\displaystyle b_{k-1}$ is odd and $\displaystyle 2b_k$ is even therefore odd+even=odd

---

You assume that the property you need to prove, i.e., that $\displaystyle b_n$ or $\displaystyle b_k$ is odd, and not the recurrence relation, holds for all previous n or k. Secondly, do you mean it is true for all n (instead of k) such that n <= k? Perhaps you could rewrite the induction step more carefully.

Hello!

Yeah I messed the things a little bit.

I meant if P(1) is true and for every integer k>=1 P(1), P(2), ... , P(k) imply P(k+1) then P(n) is true for all n>=1

Is this ok?

Quote:

---

I meant if P(1) is true and for every integer k>=1 P(1), P(2), ... , P(k) imply P(k+1) then P(n) is true for all n>=1

---

This is correct.