# Thread: What's wrong with my algebra here?

1. ## What's wrong with my algebra here?

Okay, I'm not gonna detail this summation proof, but this is what the inductive step comes down to:

I have to prove that $\displaystyle (k^2-1) + (2(k+1) +1)$ is the same as $\displaystyle (k+1)^2 -1$

$\displaystyle (k^2-1) + (2(k+1) +1)$
$\displaystyle =k^2 -1 + 2k + 2 + 1$
$\displaystyle = k^2 + 2k + 2$

Since this can't factor, I took made it look like this:
$\displaystyle =[k^2 + 2k + 1] +1$
$\displaystyle = [(k+1)(k+1)]+1$
$\displaystyle = (k+1)^2 + 1$

So I can't seem to prove that it's $\displaystyle (k+1)^2-1$, but the +1 algebra works...

In order for this proof to work, it has to be -1, anyone see anything wrong with what I did?

Thanks.

2. Those two are not the same. For example, plug in $\displaystyle k=1$.

We have: $\displaystyle (1^2-1) + (2(1+1) +1) = 5$

But: $\displaystyle (1+1)^2 - 1 = 3$

You sure that it shouldn't be: $\displaystyle (k+1)^2 +1$?

Post the question in its whole so maybe we can track down what's wrong.

3. Originally Posted by o_O
Those two are not the same. For example, plug in $\displaystyle k=1$.

We have: $\displaystyle (1^2-1) + (2(1+1) +1) = 5$

But: $\displaystyle (1+1)^2 - 1 = 3$

You sure that it shouldn't be: $\displaystyle (k+1)^2 +1$?

Post the question in its whole so maybe we can track down what's wrong.
Oh wow, thanks. Didn't notice that.

The problem was my formula was one off from the summation which made the difference. So fixing it, the new inductive step was to prove:

$\displaystyle ((k+1)^2-1) + (2(k+1)+1) = (k+2)^2-1$

which works out perfectly with algebra.

Thanks a lot!