Need help with a generalized formula for factoring

[Jacob Chestnut]

Hello,

I’ve come upon a problem in my transitional mathematics course, wherein I need to prove a generalized formula for the factorization of X^n-Y^n where the first term is
(X-Y). I have deduced a formula making use of the summation of X*Y over the range of positive integers ending at n, but this formula seems over complicated and hard to work with in an inductive proof. I’d post my solution but I’m unable to use the mathematics display software that I see some people using.

I’ve taken a look on google and I can’t seem to find any mention of this general formula, so I’d like it if someone could point me to the standard formula so I can check my work before getting into my proof.

Thanks in advance,
Jacob

 

[Tide]

Try using long division of the polynomial by x - y.

 

[Jacob Chestnut]

Thanks for the advice, but that’s not really what I need to know.

It’s really easy to figure out what the second term is for a specific value of n, but I’m trying to find a general formula for a general value of n. The pattern is even easy to see, but I want to know if anyone knows of a general standard formula for the second term as a summation from 1 to n.

 

[Tide]

Thanks for the advice, but that’s not really what I need to know.

It’s really easy to figure out what the second term is for a specific value of n, but I’m trying to find a general formula for a general value of n. The pattern is even easy to see, but I want to know if anyone knows of a general standard formula for the second term as a summation from 1 to n.

But you can do the long division for general n! You will find that the second factor (after dividing by x -y) is a geometric series.

 

[Jacob Chestnut]

I'm sorry; I’m not familiar with a geometric series in two variables. Would the multiplicative factor in this case be x^(-1)*y?

Thanks for the help,
Jacob

 

[Tide]

[tex]x^n - y^n = (x-y)\left(x^n + x^{n-1}y + x^{n-2}y^2 + \cdot \cdot \cdot + x^2 y^{n-2} + x y^{n-1} + y^n\right)[/tex]