# [SOLVED]a^n - b^n

#### dwsmith

##### Well-known member
$n\in\mathbb{N}$, prove
$$a^n-b^n = (a-b)\sum\limits_{k = 0}^{n-1}a^kb^{n-k-1}.$$

To show this is it best to just divide $a^n-b^n$ by $a-b$, show that polynomial is the summation, and then show that $(a-b)$ times the sum is $a^n-b^n$?

Or is there a more efficient method?

#### Sudharaka

##### Well-known member
MHB Math Helper
$n\in\mathbb{N}$, prove
$$a^n-b^n = (a-b)\sum\limits_{k = 0}^{n-1}a^kb^{n-k-1}.$$

To show this is it best to just divide $a^n-b^n$ by $a-b$, show that polynomial is the summation, and then show that $(a-b)$ times the sum is $a^n-b^n$?

Or is there a more efficient method?
Hi dwsmith,

You can use Mathematical induction for this one.

Kind Regards,
Sudharaka.

#### CaptainBlack

##### Well-known member
$n\in\mathbb{N}$, prove
$$a^n-b^n = (a-b)\sum\limits_{k = 0}^{n-1}a^kb^{n-k-1}.$$

To show this is it best to just divide $a^n-b^n$ by $a-b$, show that polynomial is the summation, and then show that $(a-b)$ times the sum is $a^n-b^n$?

Or is there a more efficient method?
\begin{aligned}\frac{(a-b)\sum\limits_{k = 0}^{n-1}a^kb^{n-k-1}}{b^n}\ & = (a/b-1)\sum\limits_{k = 0}^{n-1}(a/b)^k \\ &= (a/b)^n-1\end{aligned}

CB

#### dwsmith

##### Well-known member
So doing this via induction.

Let $p(n): a^n-b^n = (a-b)\sum\limits_{k=0}^{n-1}a^kb^{n-1-k}$
Then p(1) is true.
Assume p(n) is true for a fixed but arbitrary $j\leq n$.

$$p(j): a^j-b^j=(a-b)\sum\limits_{k=0}^{j-1}a^kb^{j-1-k}$$

So the problem I am having is what is multiplied to $a^j-b^j$ to get $a^{j+1}-b^{j+1}$.

$(a^j-b^j)(a+\ldots +b)$ I can't figure out some of the other terms in the middle though.

#### Sudharaka

##### Well-known member
MHB Math Helper
So doing this via induction.

Let $p(n): a^n-b^n = (a-b)\sum\limits_{k=0}^{n-1}a^kb^{n-1-k}$
Then p(1) is true.
Assume p(n) is true for a fixed but arbitrary $j\leq n$.

$$p(j): a^j-b^j=(a-b)\sum\limits_{k=0}^{j-1}a^kb^{j-1-k}$$

So the problem I am having is what is multiplied to $a^j-b^j$ to get $a^{j+1}-b^{j+1}$.

$(a^j-b^j)(a+\ldots +b)$ I can't figure out some of the other terms in the middle though.
Clearly the statement is true for $$n=1$$. Let us assume that the statement is true for $$n=p\in\mathbb{Z}^{+}$$. Then,

$a^p-b^p = (a-b)\sum_{k=0}^{p-1}a^k b^{p-1-k}$

$\Rightarrow \sum_{k=0}^{p-1}a^k b^{p-1-k}=\frac{a^p-b^p}{a-b}$

Now consider the case when $$n=p+1$$. That is, $$\displaystyle\sum_{k=0}^{p}a^k b^{p-k}$$.

\begin{eqnarray}

\sum_{k=0}^{p}a^k b^{p-k}&=&a^p+\sum_{k=0}^{p-1}a^k b^{p-k}\\

&=&a^p+b\sum_{k=0}^{p-1}a^k b^{p-k-1}\\

&=&a^p+\frac{b(a^p-b^p)}{a-b}\\

&=&\frac{a^{p+1}-ba^p+ba^p-b^{p+1}}{a-b}\\

&=&\frac{a^{p+1}-b^{p+1}}{a-b}\\

\therefore (a-b)\sum_{k=0}^{p}a^k b^{p-k}=a^{p+1}-b^{p+1}

\end{eqnarray}

Therefore by mathematical induction we have,

$a^n-b^n = (a-b)\sum\limits_{k = 0}^{n-1}a^kb^{n-k-1}\mbox{ for }n\in\mathbb{Z}^{+}$

Kind Regards,
Sudharaka.