Taking my first tottering steps with induction here. Of course, the math lecture tomorrow will handle this, but I like to keep a little bit ahead so I hope it's okay to post a thread here and ask if I've understood the fundamentals!
Now, I've spent the last hour reading the chapter and done this exercise, so basically, if someone could just point out any errors I'd be grateful:

"Prove by induction that the sum of all the n first uneven integers equals n^2."

Solution:
The sum of the first n integers = 1 + 3 + 5 +...+(2n-1) --> n(2n-1)

Assume that n(2n-1) = n^2 for all n >= 1

I: For n = 1 we have 1*(2*1-1) = 1 and 1^2 = 1
The assumption is true for n = 1

II: Assume that the statement is true for n = x
If so, then
(x+1)(2*(x+1)-1) = (x+1)^2

Proof: (x+1)(2*(x+1)-1) = (x+1)^2 QED

III: According to the induction axiom, it follows from I and II that
n(2n-1) = n^2 is true for all n >= 1


A bit heavy on the formalia perhaps, but hey - freshman calculating.

__________________
"Imagination is more important than intelligence" - Albert Einstein


I've heard people say "Quit while you're ahead", hm well bugger that, I'll quit when I'm dead

No.
I believe the BS you've fallen for / trying to propagate is to prove
n^2 = n(2n-1) for all n because

(a) 1^2 = 1(2*1-1) --> OK
(b) if x^2 = x(2x-1) then (x+1)^2 = (x+1)(2x+1)

LHS = (x+1)^2 = x^2 + 2x + 1 = x(2x-1) + 2x + 1 = 2x^2 + x + 1
Not sure how you equate it to RHS

saying
(x+1)(2*(x+1)-1) = (x+1)^2
Proof: (x+1)(2*(x+1)-1) = (x+1)^2 QED
is restating the same line. Not a proof.