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.
On
If you write the right part with a sum sign, it's easier! =)
$(a−b)\sum_0^{n-1}a^kb^{n-1-k}=\sum_0^{n-1}a^{k+1}b^{n-1-k}-\sum_0^{n-1}a^kb^{n-k}$
$=\sum_1^{n}a^kb^{n-k}-\sum_0^{n-1}a^kb^{n-k}$ $=a^n-b^n$
On
using Apostol symbols
$(b-a)(b^{p-1}+b^{p-2}a+b^{p-3}a^2+ \cdots+ba^{p-2}+a^{p-1})=(b-a)\sum_{q=0}^{p-1}(b^{p-1-q}a^q)=\sum_{q=0}^{p-1}(b^{p-q}a^q)-\sum_{q=0}^{p-1}(b^{p-1-q}a^{q+1})=\sum_{q=0}^{p-1}(b^{p-q}a^q)-\sum_{q=1}^p(b^{p-q}a^q)=[b^{p-0}a^0+\sum_{q=1}^{p-1}(b^{p-q}a^q)]-[\sum_{q=1}^{p-1}(b^{p-q}a^q)+b^{p-p}a^p]=b^p1+\sum_{q=1}^{p-1}(b^{p-q}a^q)-\sum_{q=1}^{p-1}(b^{p-q}a^q)-b^0a^p=b^p+0-1a^p=b^p-a^p$
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*}