Proof for x^n-y^n=(x-y)(x^n-1+ +y^n-1)

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SUMMARY

The discussion centers on proving the identity \(x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y + \ldots + y^{n-1})\) for any integer \(n \geq 1\). The proof utilizes mathematical induction, starting with the base case \(n=1\) and assuming the statement holds for \(n=k\). The challenge arises when transitioning to \(n=k+1\), where the user struggles to simplify the expression to match the required form. A suggestion is made to manipulate the expression using the factorization involving \((x+y)\) to aid in the proof.

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Proof for x^n-y^n=(x-y)(x^n-1+...+y^n-1)

Homework Statement


The question asks to prove that for any n[itex]\geq[/itex]1,
[itex]x^{n}[/itex]-[itex]y^{n}[/itex]=(x-y)([itex]x^{n-1}[/itex]+[itex]x^{n-2}[/itex]y+...+[itex]y^{n-1}[/itex])


Homework Equations


[itex]x^{n}[/itex]-[itex]y^{n}[/itex]=(x-y)([itex]x^{n-1}[/itex]+[itex]x^{n-2}[/itex]y+...+[itex]y^{n-1}[/itex])


The Attempt at a Solution



So far, I used induction.
So for n=1, x-y=x-y

Second step, I assume that n=k is true:
[itex]x^{k}[/itex]-[itex]y^{k}[/itex]=(x-y)([itex]x^{k-1}[/itex]+[itex]x^{k-2}[/itex]y+...+[itex]y^{k-1}[/itex])

I get stuck at n=k+1.

[itex]x^{k+1}[/itex]-[itex]y^{k+1}[/itex]=(x-y)([itex]x^{k}[/itex]+[itex]x^{k-1}[/itex]y+...+[itex]y^{k}[/itex])

When I expand RHS, I get:
[itex]x^{k+1}[/itex]-[itex]x^{k}[/itex]y+[itex]x^k{}[/itex]y-[itex]x^{k-1}[/itex][itex]y^{2}[/itex]+...+x[itex]y^{k}[/itex]-[itex]y^{k+1}[/itex]

I think that I need to cancel things so I can be left only with [itex]x^{k+1}[/itex]-[itex]y^{k+1}[/itex], but I always have terms in the middle which do not cancel out.

What am I doing wrong?
 
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you need to work from the the n case to the n+1 (or vice versa)

i haven't worked it, but how about noticing:
[tex](x+y)(x^{k} -y^{k}) = x^{k+1} -xy^{k} -x^{k}y -y^{k+1}[/tex]

then you have
[tex]x^{k+1} -y^{k+1} = (x+y)(x^{k} -y^{k})-x^{k}y +xy^{k}[/tex]

then see if you can work it into the required form...
 

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