Working on homework for a probability and computing class, but my ability to work with summations is rusty to say the least, so I suspect this is going to turn out pretty straightforward.
Problem asks to find a closed formula for $$\sum_{i=1}^ni3^i$$ by representing it as a double sum and changing the order of summation. I did that by following a hint from the instructor and came up with $$\sum_{k=1}^n\sum_{i=k}^n3^i,$$ but I'm not really sure what that accomplished. What's the next step? What am I looking for here?
Answer
Here is a rather detailed elaboration which might be useful.
We obtain
\begin{align*}
\color{blue}{\sum_{i=1}^ni3^i}&=\sum_{i=1}^n\left(\sum_{k=1}^i 1\right)3^i\tag{1}\\
&=\sum_{i=1}^n\sum_{k=1}^i 3^i
=\sum_{1\leq k\leq i\leq n}3^i
=\sum_{k=1}^n\sum_{i=k}^n3^i\tag{2}\\
&=\sum_{k=1}^n\sum_{i=0}^{n-k}3^{i+k}\tag{3}\\
&=\sum_{k=1}^n3^k\cdot\frac{3^{n-k+1}-1}{3-1}\tag{4}\\
&=\frac{1}{2}\sum_{k=1}^n\left(3^{n+1}-3^k\right)\tag{5}\\
&=\frac{n}{2}3^{n+1}-\frac{1}{2}\sum_{k=1}^n3^k\tag{6}\\
&=\frac{n}{2}3^{n+1}-\frac{1}{2}\cdot\left(\frac{3^{n+1}-1}{3-1}-1\right)\tag{7}\\
&=\frac{n}{2}3^{n+1}-\frac{1}{4}3^{n+1}+\frac{3}{4}\tag{8}\\
&\color{blue}{=\frac{n}{4}(2n-1)3^{n+1}+\frac{3}{4}}\tag{9}
\end{align*}
Comment:
In (1) we represent the factor $i$ as sum.
In (2) we multiply out in the left-hand sum and write the index range somewhat more conveniently in the middle sum. We exchange the sums in the right-hand double-sum.
In (3) we shift the index of the inner sum to start from $i=0$.
In (4) we apply the finite geometric summation formula.
In (5) we do some simplifications.
In (6) we multiply out and do some simplifications.
In (7) we apply the finite geometric summation formula again.
In (8) and (9) we do some more simplifications.
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