# 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.

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Is $n=p$? If not: how are they related? – mathmax May 17 '14 at 21:53

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*}

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If you write the right part with a sum sign, it's easier! =)

$(a−b)\sum_0^{n-1}a^kb^{n-1-k}=\sum_0^{n-1}a^{k+1}b^{n-1-k}-\sum_0^{n-1}a^kb^{n-k}$

$=\sum_1^{n}a^kb^{n-k}-\sum_0^{n-1}a^kb^{n-k}$ $=a^n-b^n$

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using Apostol symbols

$(b-a)(b^{p-1}+b^{p-2}a+b^{p-3}a^2+ \cdots+ba^{p-2}+a^{p-1})=(b-a)\sum_{q=0}^{p-1}(b^{p-1-q}a^q)=\sum_{q=0}^{p-1}(b^{p-q}a^q)-\sum_{q=0}^{p-1}(b^{p-1-q}a^{q+1})=\sum_{q=0}^{p-1}(b^{p-q}a^q)-\sum_{q=1}^p(b^{p-q}a^q)=[b^{p-0}a^0+\sum_{q=1}^{p-1}(b^{p-q}a^q)]-[\sum_{q=1}^{p-1}(b^{p-q}a^q)+b^{p-p}a^p]=b^p1+\sum_{q=1}^{p-1}(b^{p-q}a^q)-\sum_{q=1}^{p-1}(b^{p-q}a^q)-b^0a^p=b^p+0-1a^p=b^p-a^p$

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