Prove by Induction: $w_k = w_{k-2} + k$

[Arew]

Homework Statement

Prove by induction $$w_k = w_{k−2} + k$$, for all integers $$k \ge 3, w_1 = 1,w_2 = 2$$ has an explicit formula
$$ w_n =\begin{cases}
\frac{(n+1)^2}{4}, & \text{if $n$ is odd} \\
\frac n2(\frac n2 + 1), & \text{if $n$ is even}
\end{cases}$$

Homework Equations

The Attempt at a Solution

[/B]
Inductive step for when n is odd:

Suppose $$w_k = \frac{(k+1)^2}{4}$$ if k is odd. Then by definition of w, we have $$w_{k + 2} = w_k + k + 2 = \frac{(k+1)^2}{4} + k + 2 = \frac {k^2 + 2k + 1}{4} + k + 2= \frac {k^2 + 6k + 8}{4} = \frac {(k +3)^2}{4} $$ if k + 2 is odd.

Is it important that we prove $$w_{k + 1} = \frac{(k+2)^2}{4}$$ if k+ 1 is odd or is the proof for $$w_{k + 2} = \frac{(k+3)^2}{4}$$ if k + 2 is enough?

If none of the above makes sense, can I please get help getting started with the inductive step.

 

[PeroK]

Homework Statement

Prove by induction $$w_k = w_{k−2} + k$$, for all integers $$k \ge 3, w_1 = 1,w_2 = 2$$ has an explicit formula
$$ w_n =\begin{cases}
\frac{(n+1)^2}{4}, & \text{if $n$ is odd} \\
\frac n2(\frac n2 + 1), & \text{if $n$ is even}
\end{cases}$$

Homework Equations

The Attempt at a Solution

[/B]
Inductive step for when n is odd:

Suppose $$w_k = \frac{(k+1)^2}{4}$$ if k is odd. Then by definition of w, we have $$w_{k + 2} = w_k + k + 2 = \frac{(k+1)^2}{4} + k + 2 = \frac {k^2 + 2k + 1}{4} + k + 2= \frac {k^2 + 6k + 8}{4} = \frac {(k +3)^2}{4} $$

It's fine to here. That's the odd numbers sorted out. Now, what about the even numbers?

 

[fresh_42]

Homework Statement

Prove by induction $$w_k = w_{k−2} + k$$, for all integers $$k \ge 3, w_1 = 1,w_2 = 2$$ has an explicit formula
$$ w_n =\begin{cases}
\frac{(n+1)^2}{4}, & \text{if $n$ is odd} \\
\frac n2(\frac n2 + 1), & \text{if $n$ is even}
\end{cases}$$

Homework Equations

The Attempt at a Solution

[/B]
Inductive step for when n is odd:

Suppose $$w_k = \frac{(k+1)^2}{4}$$ if k is odd. Then by definition of w, we have $$w_{k + 2} = w_k + k + 2 = \frac{(k+1)^2}{4} + k + 2 = \frac {k^2 + 2k + 1}{4} + k + 2= \frac {k^2 + 6k + 8}{4} = \frac {(k +3)^2}{4} $$ if k + 2 is odd.

Is it important that we prove $$w_{k + 1} = \frac{(k+2)^2}{4}$$ if k+ 1 is odd or is the proof for $$w_{k + 2} = \frac{(k+3)^2}{4}$$ if k + 2 is enough?

If none of the above makes sense, can I please get help getting started with the inductive step.

And the induction base? You have to verify this, too. It is really important and not just something annoying.
And what if k is even?
(Remark: there is a typo.)

 

[Arew]

It's fine to here. That's the odd numbers sorted out. Now, what about the even numbers?

And the induction base? You have to verify this, too. It is really important and not just something annoying.
And what if k is even?
(Remark: there is a typo.)

Thanks for the comments. Do we have to handle $$w_{k+1}$$ at all?

 

[PeroK]

Thanks for the comments. Do we have to handle $$w_{k+1}$$ at all?

Why would you?

 

[Arew]

Why would you?

I was worried we proved w_k for every other integer k. I see why that's wrong. Thanks.

 

[fresh_42]

I was worried we proved w_k for every other integer k. I see why that's wrong. Thanks.

With the induction base (two values), the odd numbers, which are covered by $k \rightarrow k+2$, only the even are missing. What other numbers can you think of?

 

[Arew]

With the induction base (two values), the odd numbers, which are covered by $k \rightarrow k+2$, only the even are missing. What other numbers can you think of?

Can't think of anything else :) Thanks.