Using telescoping property to prove difference of powers Ok so I have started working through Apostol calculus and as you can see I am stuck.
The problem is that I can not see the telescoping pattern anywhere for following problem.
Prove that $$a^n - b^n = (a-b)(a^{n-1} + a^{n-2}b + ... +  ab^{p-2} + b^{p-1})$$
using telescoping propery
Usage of telescoping property is actually a hint to the problem but it actually made my life much harder.
Any ideas?I would be very thankful for swift and quick answer.
 A: You probably have some typos in your statement. Once you get the correct identity to prove, it's a few lines to finish the proof. Open the parentheses of your right-hand-side
\begin{align*}
(a-b)\sum_{k=1}^{n} a^{n-k} b^k &= (a-b)(a^{n-1} + a^{n-2}b^1 + \ldots + a^1 b^{n-2} + b^{n-1})\\
&= a^n + a^{n-1}b^1 + \ldots + a^2 b^{n-2} + a^1 b^{n-1}\\
& \quad - (a^{n-1}b^1 + a^{n-2}b^2 + \ldots + a^1 b^{n-1} + b^{n})\\
& = a^n - b^n
\end{align*}
A: If you write the right part with a sum sign, it's easier! =)
$(a−b)\sum_0^{n-1}a^kb^{n-1-k}=\sum_0^{n-1}a^{k+1}b^{n-1-k}-\sum_0^{n-1}a^kb^{n-k}$
$=\sum_1^{n}a^kb^{n-k}-\sum_0^{n-1}a^kb^{n-k}$
$=a^n-b^n$
A: using Apostol symbols
$(b-a)(b^{p-1}+b^{p-2}a+b^{p-3}a^2+ \cdots+ba^{p-2}+a^{p-1})=(b-a)\sum_{q=0}^{p-1}(b^{p-1-q}a^q)=\sum_{q=0}^{p-1}(b^{p-q}a^q)-\sum_{q=0}^{p-1}(b^{p-1-q}a^{q+1})=\sum_{q=0}^{p-1}(b^{p-q}a^q)-\sum_{q=1}^p(b^{p-q}a^q)=[b^{p-0}a^0+\sum_{q=1}^{p-1}(b^{p-q}a^q)]-[\sum_{q=1}^{p-1}(b^{p-q}a^q)+b^{p-p}a^p]=b^p1+\sum_{q=1}^{p-1}(b^{p-q}a^q)-\sum_{q=1}^{p-1}(b^{p-q}a^q)-b^0a^p=b^p+0-1a^p=b^p-a^p$
