Summation/Sigma Notation Question This is my first time on here.  I am in first year engineering, and I'm having some trouble with sigma notation.  Here is the question and answer:

I am trying to convert the summation into closed form, using the rules of summation, and the summation formulas.  I was able to get the same answer, except that my answer did not multiply the first term by pi.  Is there anyone able to explain how to do this, or give me some insight so that I might try it again in a different way?
Thank you so much.  Your time is greatly appreciated.
 A: First split the sum
\begin{align}\sum\limits_{k=1}^n (\pi^k-3) &= \sum\limits_{k=1}^n\pi^k -\sum\limits_{k=1}^n 3\\ &= \sum\limits_{k=1}^n\pi^k -3\sum\limits_{k=1}^n 1\\
&= \sum\limits_{k=0}^{n-1}\pi^{k+1} -3\sum\limits_{k=1}^n 1\\ 
&= \pi\sum\limits_{k=0}^{n-1}\pi^{k} -3n,\end{align}
where I shifted the index in the first sum. This first sum is a geometric series (with $a = 1$ and $r = \pi$). So we get
$$\sum\limits_{k=1}^n (\pi^k-3) = \pi\frac{1-\pi^n}{1-\pi}-3n = \frac{\pi (\pi^n-1)}{\pi-1}-3n.$$
A: $$\begin{align}
\sum_{k=1}^n(\pi^k-3) & = \sum_{k=1}^n\pi^k-3\sum_{k-1}^n 1\\
& = \pi\sum_{k=1}^n\pi^{k-1}-3n\\
& = \pi\sum_{k=0}^{n-1}\pi^k-3n\\
& =\pi{\pi^n-1\over\pi-1}-3n.
\end{align})$$
A: \begin{align} \sum_{k=1}^n (\pi^k - 3) &= -3 n + \sum_{k=1}^n \pi^k = - 3n + (\pi + \pi^2 + \cdots + \pi^n) = \\  &= - 3n + \frac{(\pi - 1)(\pi + \pi^2 + \cdots + \pi^n)}{\pi - 1} =\\ &=- 3n + \frac{\pi^2 + \pi^3 + \cdots + \pi^n + \pi^{n+1} - \pi - \pi^2 - \pi^3 \cdots - \pi^n }{\pi - 1} =\\ &= -3n +\frac{\pi^{n+1} - \pi}{\pi - 1} = \frac{\pi(\pi^{n} - 1)}{\pi - 1}-3n \, .
\end{align}
