1. ## Oh painful induction!!!

Well, here's another topic I'm horrible at: Induction.

I think I can do bulk-standard induction, but just to make sure tell me if the following example is correct:

1 + 3 + 5 + 7 + .... + (2n -1) = n2, ε Z+

First let us prove this proposition P works for 1:

P(1) = 2(1) - 1 =1

This must equal n2 = 1 .: P(1) is true and works

Now we assume the proposition works for k:

P(k) = K 2

P(K +1) must therefore equal: K2 + (2(K+1) -1) = 2K + 2 -1 = 2K + 1

P(K+1) = (K +1) (K + 1) = K2 + 2K + 1...

Therefore if P(k) is true then P(K+1) is always true proving the propision is true.

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I'm hoping I got this right. Tell me if I did...

But then how do we do induction with matrices, calculus, geommetric series etc. ??

Can anyone teach me in a easy way (showing all steps) cause I'm as dumb as a bag of rocks!!!

I have to urgently know this for a test, and the textbook just looks too confusing for me! Please try to make it as simple and easy to udnerstand as possible! Thank you!

- Joey

2. ## Induction formatting

Again I'm having trouble with the formatting, all the squared and powers formatting have been removed. All numbers after letter should be powers. If you'd like the nicely formatted version of this thread, I've attached a word document which shows it correctly.

Thank you for your time and patience.

3. For your inductive step assume that P(k) is true. So we know

$1+3+5+ \dots+ 2k-1=k^2$

and we want to show P(k+1) is true.

Therefore (by adding 2k+1 to both sides)

$1+3+5 +\dots +2k-1+2k+1=k^2 +2k+1=(k+1)^2$.

Can you drive from here? ( You might want to note that 2k+1=2(k+1)-1

4. For your inductive step assume that P(k) is true. So we know

and we want to show P(k+1) is true.

Therefore (by adding 2k+1 to both sides)

.

Can you drive from here? ( You might want to note that 2k+1=2(k+1)-1
LoL sorry I'm confused. Read the document above.

We have from our hypothesis that

$1+3+5+\cdots+2n+1=n^2~\forall{n}\in\mathbb{N}$

So now we must test this for the $n+1$th term. Setting this up we get

$+3+5+\cdots+2(n-1)+1+2(n+1)+1=(n+1)^2=n^2+2n+1$

Doing a little algebra on the last two terms of the left hand side gives

$1+3+5+\cdots+2n+1+2n+1=n^2+2n+1$

Now from our hypothesis $n^2=1+3+5+\cdots+2n+1$ so using that fact we rewrite this as

$1+3+5+\cdots+2n+1+2n+1=\left(1+3+5+\cdots+2n+1\rig ht)+2n+1$

So now subtracting $1+3+5+\cdots+2n+1$ gives

$2n+1=2n+1$, so we have proved the inductive step.