Question

Could someone review this proof for me?

Prove:
A^n - B^n = (A-B)×(A+B)^n-1

(A+B)^n-1 = (A^n-1)+(A^n-2)(B^1)+(A^n-3)(B^2)+...+(A^1)(B^n-2)+(B^n-1)

(A-B)(A+B)^n-1 = [(A^n)+(A^n-1)(B^1)+(A^n-2)(B^2)+...+(A^2)(B^n-2)+(A^1)(B^n-1)] - [(A^n-1)(B^1)+(A^n-2)(B^2)+(A^n-3)(B^3)+...+(A^1)(B^n-1)+(B^n)]

Cancel out like terms:
(A-B)(A+B)^n-1= (A^n)-(B^n)

Answer 1

Prove:
\[A^n - B^n = (A-B)(A+B)^{ n-1}\]
Binomial exponents:
\[(A+B)^{n-1} = (A^{n-1})+(A^{n-2})(B^1)+(A^{n-3})(B^2)+...+(A^1)(B^{n-2})+(B^{n-1})\]
Factoring in (A-B):
\[(A-B)(A+B)^{n-1} = [(A^n)+(A^{n-1})(B^1)+(A^{n-2})(B^2)+...+(A^2)(B^{n-2})+(A^1)(B^{n-1})]\]
\[- [(A^{n-1})(B^1)+(A^{n-2})(B^2)+(A^{n-3})(B^3)+...+(A^1)(B^{n-1})+(B^n)]\]
Cancelling like terms:
\[(A-B)(A+B)^{n-1}= A^n-B^n\]

Answer 2

I think it's right but I would like someone else to make sure. I'd ask one of my proffs but I won't be seeing any of them for another 2 months.

Answer 3

Do you know how to prove things using induction?

Answer 4

Is that the "Prove it for n=1" then "Assume it's true for n and prove it for n+1"?

Answer 5

Yup, that's it.

Answer 6

I don't know what kind of rigor is required here, but that's how I'd go about proving this.

Answer 7

the (n+1) case will yield a part of the original base case. My assumption is that when going through that step you'll wide up with a bunch of stuff that'll cancel out and arrive at the conclusion that they are indeed equal.

Answer 8

Cool, thanks!

Answer 9

No problem! Induction is really such an awesome tool. :D