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This question is from [Number Theory George E. Andrews 1-1 #3].

Prove that $$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}).$$

This problem is driving me crazy.
$$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\dots +xy^{n-2}+y^{n-1)}$$

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

This is a similar problem in the book and I tried this method but it wasn't working out

$\quad$Thereom $\bf1$-$\bf2$: $\,\,\,\,$ If $\,x$ is any real number other than $1$, then $$\sum_{j=0}^{n-1}x^j=1+x+x^2+\ldots+x^{n-1}=\dfrac{x^n-1}{x-1}.$$ $\quad$Remark: $\displaystyle\sum_{j=0}^{n-1}A_j$ is shorthand for $A_0+A_1+A_2+\ldots+A_{n-1}.$
$\quad$Proof: Again we proceed by mathematical induction. If $n=1$ then $\displaystyle\sum_{j=0}^{1-1}x^j=x^0=1$ and $(x-1)/(x-1)=1$. Thus the theorem is true for $n=1$.
$\quad$ Assuming that $\displaystyle\sum_{j=0}^{k-1}x^j=(x^k-1)/(x-1)$, we find that $$ \eqalign{ \sum^{(k+1)-1}_ {j=0}x^j & = \sum^{k-1}_ {j=0}x^j+x^k=\dfrac{x^k-1}{x-1}+x^k \\ &= \dfrac{x^k-1+x^{k+1}-x^k}{x-1}\\ &= \dfrac{x^{k+1}-1}{x-1}. }$$ Hence condition $(\rm ii)$ is fulfilled, and we have established the theorem.

$\quad$Corollary $\bf1$-$\bf1$: $\,\,$ If $\,m$ and $n$ are positive integers and if $m>1$, then $n<m^n.$

Martin Sleziak
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O.rka
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  • The second image is not a problem, but rather a theorem with a complete proof. Note that the first problem can be solved with this theorem by plugging in $\frac xy$ for §x§ and multiplying with $y^n$ (what care should be taken if $y=0$?) – Hagen von Eitzen Jan 20 '13 at 22:14

4 Answers4

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Hint: Apply $\textbf{Theorem 1.2}$ to $\displaystyle \frac{x}{y}$.

Edit: It follows from $\textbf{Theorem 1.2}$ that $\displaystyle \Bigl(\frac{x}{y}\Bigr)^n -1=\Bigl(\frac{x}{y}-1\Bigr)\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$.

Now multiply the equation by $y^n$ to get

$$\displaystyle y^n\Bigl(\Bigl(\frac{x}{y}\Bigr)^n -1)\Bigr)=y^n\Bigl(\frac{x}{y}-1\Bigr)\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$$

Simplifying on the left-hand side and rewritting $y^n$ as $yy^{n-1}$ on the right-hand side we get

$$(x^n -y^n)=yy^{n-1}\Bigl(\frac{x}{y}-1\Bigr)\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$$

Because the product is commutative you can rewrite the right-hand side to get

$$(x^n -y^n)=y\Bigl(\frac{x}{y}-1\Bigr)y^{n-1}\Bigl( \Bigl(\frac{x}{y}\Bigr)^{n-1}+\cdots +\frac{x}{y}+1\Bigr)$$

Finally, on the right-hand side, factor in $y$ and $y^{n-1}$ accordingly to get

$$(x^n -y^n)=(x-y)(x^{n-1}+\cdots +xy^{n-2}+y^{n-1})$$

Git Gud
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  • So I tried expanding it if it were (x/y) – O.rka Jan 20 '13 at 22:56
  • @draconisthe0ry was that a question somehow? – Git Gud Jan 20 '13 at 23:23
  • Sorry about the last comment . . I didn't know that I ended up pressing the Add Comment button. I see that this works but what confuses me is that when you divide both sides by (x-y) and you get (x^(n−1)+x^(n−2)y+⋯+xy^(n−2)+y^(n−1)) It looks like "y" is following the theorem in the sense that it propagates as (y^0+y^1+...+y^(n-2)+y^(n-1)) and it is being multiplied that same sequence in reverse for x. I don't understand how the sequence in reverse for x is being multiplied in reverse order. – O.rka Jan 22 '13 at 01:44
  • @draconisthe0ry I rewrote my answer with a more detailed explanation. Can you understand it now? – Git Gud Jan 22 '13 at 07:24
  • yes this answer is perfect. i understand each step. the only concern I have is how to think of initially using (x/y) instead of x in Theorem 1-2. It makes sense on how that works since you can pull a y^n/y out of (x^n-y^n)/(x-y) to get ((x/y)^n-1)/(x/y-1) . Does this prove that the expression works for n+1 ? To prove it works for n+1 as well I would just show that ((x/y)^n-1)/(x/y-1) + (x/y)^n = ((x/y)^(n+1)-1)/((x/y)-1) right ? – O.rka Jan 22 '13 at 17:43
  • @draconisthe0ry It's a trick. You got to have the inspiration to find it or already know it. Regarding your question, this was a proof for all $n\in \mathbb{N}$, but I didn't use induction. You don't need it. If you wanna prove it by induction either take a look at Brian's answer or use the same idea in the induction step, but you won't have to use the induction hypothesis if you do that. – Git Gud Jan 22 '13 at 17:56
  • @draconisthe0ry On my comment above I meant "take a look at Brian's answer or use the same idea I USED in my answer". – Git Gud Jan 22 '13 at 18:09
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Your induction hypothesis is that

$$\sum_{j=0}^{k-1}x^jy^{(k-1)-j}=\frac{x^k-y^k}{x-y}\;.$$

Now follow the model:

$$\begin{align*} \sum_{j=0}^{(k+1)-1}x^jy^{k-j}&\overset{(1)}=\left(\sum_{j=0}^{k-1}x^jy^{k-j}\right)+x^ky^0\\ &\overset{(2)}=y\left(\sum_{j=0}^{k-1}x^jy^{(k-1)-j}\right)+x^k\\ &\overset{(3)}=y\cdot\frac{x^k-y^k}{x-y}+x^k\\ &\overset{(4)}=\frac{x^ky-y^{k+1}+x^{k+1}-x^ky}{x-y}\\ &\overset{(5)}=\frac{x^{k+1}-y^{k+1}}{x-y}\;. \end{align*}$$

$(1)$ is splitting off the last term of the summation; $(2)$ factors a $y$ out of the remaining summation; $(3)$ uses the induction hypothesis; and $(4)$ and $(5)$ are just algebra.

Brian M. Scott
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  • I'm trying to make sense of it now . . . So I see that (1 + y^1 + ... + y^(n-1)) = (y^n-1)/(y-1) but it is being multiplied by (x^(n-1)+...+x^1+1) so that is equal to (x^n-1)/(x-1) but would that work when you multiply them ? – O.rka Jan 20 '13 at 22:47
  • @draconisthe0ry: I’m not sure what you’re asking. I’m simply observing that $$x^n-y^n = (x-y)(x^{n-1}+x^{n-2}y+\dots+xy^{n-2}+y^{n-1)};,$$ the statement that you want to prove, is equivalent to $$\sum_{k=0}^{n-1}x^ky^{(n-1)-k}=\frac{x^n-y^n}{x-y}$$ and then proving the latter by induction. (Actually, they’re not equivalent when $x=y$, but in that case the first one is trivially true, since both sides are $0$.) This is exactly what your text did with the $x^n-1$ identity: it converted $x^n-1=(x-1)(\text{stuff})$ to $\text{stuff}=\frac{x^n-1}{x-1}$ and proved that. – Brian M. Scott Jan 20 '13 at 22:56
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$$\frac{1-(x/y)^n}{1-x/y}=1+x/y+(x/y)^2+...+(x/y)^{n-1}$$ $$\frac{(y^n-x^n)/y^n}{(y-x)/y}=\frac{y^{n-1}+xy^{n-2}+...+x^{n-1}}{y^{n-1}}$$ $$\frac{y^n-x^n}{(y-x)y^{n-1}}=\frac{y^{n-1}+xy^{n-2}+...+x^{n-1}}{y^{n-1}}$$ $$y^n-x^n=(y-x)(y^{n-1}+xy^{n-2}+...+x^{n-1})$$

Adi Dani
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    Effective answer. +1 – Mikasa Jan 21 '13 at 10:32
  • Ok so I see how you switched x and y . . . I also see how you can pull out y^n from the numerator and y from the denominator to get ((x/y)^n-1)/((x/y)-1) . what i don't understand is when you multiply the top and bottom of the RHS by y^(n-1) you get y^(n-1)+y^(n-2)+...+y^0 ? I get why the first two are y^(n-1) and y^(n-2) but as the sequence progresses wouldn't it keep going to y^(n-infinity) . i feel like that is a dumb question but i'm not getting it and that's the one part that is stopping me from finishing the problem – O.rka Jan 22 '13 at 04:10
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Let $u_n$ $=$ $x^n-y^n$. Now note that $u_n$ $=$ $(x+y)u_n$$_-$$_1$ $+$ $xy$ $u_n$$_-$$_2$. Assume that the given expression is true for all numbers from $1$ to some fixed $k(>1)$. Then apply induction to prove the result for $(k+1)$.

William Hilbert
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