# Induction Proof Help

1. Jan 20, 2013

### hicsuntdrac0ni

1. The problem statement, all variables and given/known data

https://sphotos-b.xx.fbcdn.net/hphotos-prn1/69668_10151632316928154_624610826_n.jpg [Broken]

x^n-y^n = (x-y)*(x^(n-1)+x^(n-2)*y+...+x*y^(n-2)+y^(n-1))

from Number Theory by George E. Andrews
2. Relevant equations

3. The attempt at a solution

(x^n-y^n)/(x-y) = the sum for the first n numbers and then i added (x*y^((n+1)-2)+y^((n+1)-1)) which should equal (x^(n+1)-y^(n+1))/(x-y) but i can't figure it out

Last edited by a moderator: May 6, 2017
2. Jan 20, 2013

### vela

Staff Emeritus
That's not right. If you let $n \rightarrow n+1$, you get
\begin{align*}
x^{n+1}-y^{n+1} &= (x-y)(x^{(n+1)-1}+x^{(n+1)-2}y+\cdots+xy^{(n+1)-2}+y^{(n+1)-1}) \\
&= (x-y)(x^n+x^{n-1}y+\cdots+xy^{n-1}+y^{n})
\end{align*} You tacked on the last two terms, but all of the other terms in the sum don't match.

3. Jan 20, 2013

### hicsuntdrac0ni

The expressions after the "+...+" are supposed to be the nth term so the n+1 term should be those last two with (n+1) substituted for (n) . Since the first nth terms are x^(n-1)+x^(n-2)*y+...+x*y^(n-2)+y^(n-1) and that is equal to x^n-y^n/(x-y) then I can substitute the x^(n-1)+x^(n-2)*y+...+x*y^(n-2)+y^(n-1) for x^n-y^n/(x-y) and then add x^(n+1)-y^(n+1)/(x-y)

4. Jan 20, 2013

### vela

Staff Emeritus
I just showed you the first n-1 terms in the sum aren't what you think they're equal to.