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Let $x,y \in \mathbb{R}$.

Show that $$ x^{n+1}-y^{n+1}=(x-y)\sum_{k=0}^n x^k y^{n-k} $$

for all $n \in\mathbb{N}_0$

I need to prove this via induction.

My attempt: base case (k=0) = $$x^{0+1}-y^{0+1}=(x-y)(x^0 y^{n-0}) $$ $$x-y=(x-y)(y^n)$$ Here is where I get lost, does this disprove this? This statement is only true if $y^n = 1$? Doesn't this mean that this isn't true for all $x,y$ in $\mathbb R$?

Martin Sleziak
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Mathissohardlmao
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    for the base case we have $x^{0+1}-y^{0+1}=(x-y)(x^0y^0)=(x-y)1=x-y,$ which is true; are you getting confused because you're using $k$ for both the index variable in the sum and also the case of the induction? – J. W. Tanner Sep 17 '19 at 03:23
  • Note that you have mixed up your notation here, your $y^n$ should be $y^k$, after all you are doing induction over $k$, not $n$. – koifish Sep 17 '19 at 03:26
  • @J.W.Tanner I'm totally confused. I don't know whether the n in $y^n$ is a constant or not and whether that changes when I take the inductive step. Also why do you say (x-y)1=x-y? Where does the $y^n$ go – Mathissohardlmao Sep 17 '19 at 03:31
  • $n$ in $y^n$ is like $n$ in $x^{n+1}$; maybe you would be less confused if you used a letter other than $k$ (such as $j$) for your summation variable – J. W. Tanner Sep 17 '19 at 03:33
  • The induction variable is $n$; $k$ is just a "dummy" variable. The first few cases ($n=0$, $1$, $2$) are $$x-y=(x-y)1,$$ $$x^2-y^2=(x-y)(x+y),$$ $$x^3-y^3=(x-y)(x^2+xy+y^2).$$ – Angina Seng Sep 17 '19 at 03:34

2 Answers2

1

The equation you're asking to prove is

$$x^{n+1}-y^{n+1}=(x-y)\sum_{k=0}^n x^k y^{n-k} \tag{1}\label{eq1}$$

for all $n \in \mathbb{N_{0}}$. As mentioned in multiple question comments and in Trevor Gunn's answer, the base case of $n = 0$ is true (as you use induction on the limit variable, not the summation one of $k$) since $x - y = (x - y)x^0 y^0$ and $x^0 = y^0 = 1$.

Continuing the induction proof, assume \eqref{eq1} is true for $n = m$ for some $m \ge 0$, i.e.,

$$x^{m+1}-y^{m+1}=(x-y)\sum_{k=0}^m x^k y^{m-k} \tag{2}\label{eq2}$$

The RHS of \eqref{eq1} for $n = m + 1$ is

$$\begin{equation}\begin{aligned} (x-y)\sum_{k=0}^{m+1} x^k y^{m+1-k} & = (x-y)\left(\sum_{k=0}^{m} x^k y^{m+1-k} + x^{m+1}y^{0}\right) \\ & = (x-y)\left(y\sum_{k=0}^{m} x^k y^{m-k} + x^{m+1}\right) \\ & = y\left((x-y)\sum_{k=0}^{m} x^k y^{m-k}\right) + (x-y)x^{m+1} \\ & = y\left(x^{m+1}-y^{m+1}\right) + x^{m+2} - yx^{m+1} \\ & = yx^{m+1} - y^{m+2} + x^{m+2} - yx^{m+1} \\ & = x^{m+2} - y^{m+2} \end{aligned}\end{equation}\tag{3}\label{eq3}$$

This is the LHS of \eqref{eq1} for $n = m + 1$. This shows \eqref{eq1} is true for $n = m + 1$ if it's true for $n = m$, and since it's true for $n = 0$, induction shows \eqref{eq1} is true for all $n \in \mathbb{N_{0}}$.

John Omielan
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When you apply the principle of induction to an identity involving a sum, the base case and induction step is always in terms of the variable on the top of the summation ($n$) and not the indexing variable ($k$). So the base case is $n = 0$ which looks like

$$x^{0 + 1} - y^{0 + 1} = (x - y)\sum_{k = 0}^0x^ky^{0-k} = (x - y)(x^0y^0). $$

Trevor Gunn
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