Proving an algebraic identity for all odd/even integers

Question

Trying to prove these two different facts, firstly that

$\forall$ odd $n \in \mathbb{N}$,
$x^n + y^n = (x+y) \sum_{i=0}^{n-1} (-1)^ix^{n-1-i}y^i$

and then that

$\forall$ even $n \in \mathbb{N}$,
$x^n - y^n = (x+y) \sum_{i=0}^{n-1} (-1)^ix^{n-1-i}y^i$

I know that we have that for $x, y \in \mathbb{R}$, and $n \in \mathbb{Z}^{+}, x^n -y^n = (x-y)\sum_{i=0}^{n-1} x^{n-1-i}y^i$.

Unsure about the initial approach. Thinking about it, I know in the case where n is odd, if we write out the summation, we get $x^{n-1} - x^{n-2} y +\ ...\ - y^{n-2} + y^{n-1}.$ When n is even, instead we have $x^{n-1} - x^{n-2} y +\ ...\ + y^{n-2} - y^{n-1}.$ I'm guessing there's some cancelling out to do from here but not too sure.

Or maybe a proof by induction would be easier?

Answer 1

Hint:

The simplest way, in my opinion, is to prove first the non-homogeneous equivalent formulæ: \begin{align} &1+t^{2m+1}=(1+t)(1-t+t^2-\dots+t^{2m}) \\ &1-t^{2m}=(1+t)(1-t+t^2-\dots-t^{2m-1}) \end{align}

The induction for the first should then be easy. You also can expand the r.h.s. on two rows and see clearly the telescoping sum.

For the second formula, factor the l.h.s. as $$1-(t^2)^m=(1-t^2)\bigl(1+t^2+\dots +t^{2(m-1)}\bigr)=(1+t)(1-t)\bigl(1+t^2+\dots +t^{2m-2)}\bigr),$$ and expand the product of the last two factors on two rows, again.

Final step: for the homegeneous case, set $t=\dfrac yx$ and the binomials as $$x^n+y^n=x^n(1+t^n),\qquad x^n-y^n=x^n(1-t^n) ,$$ apply the above formulæ and re-homogeneise.

Answer 2

$$a^{k+1}-a^k=(a-1)a^k $$

Apply the sum $\sum\limits^{n-1}_{k=0}$ in both sides, then by telescopic sum we have

$$a^{n}-1=(a-1) \sum^{n-1}_{k=0}a^k$$

take $a=\frac{x}{y}$, so

$$ \frac{x^n-y^n}{y^n}=(x-y)\sum^{n-1}_{k=0}x^{k}y^{-k-1}$$

then $$ x^n-y^n=(x-y)\sum^{n-1}_{k=0}x^{k}y^{n-k-1}=$$ by symmetry $$=(x-y)\sum^{n-1}_{k=0}x^{n-k-1}y^{k}.$$

Iff $n$ is odd, then $(-1)^n=-1$. So in the place of y take -y, with n odd

$$ x^n+y^n=(x+y)\sum^{n-1}_{k=0}(-1)^{k}x^{n-k-1}y^{k}.$$