Convert this sum to normal expression

Question

If I have a sum like this: $$2\sum_{i=0}^{n-1}3^i(3^{n-i}-1)$$ How do I convert it so that I can lose the sum. For example if it was $$\sum_{i=0}^{n}n$$ then the result would be $$\frac{n(n-1)}{2}$$ Is there a general principle how to do this, for example like for geometric sums?

Answer 1

The general principles are the rules of arithmetic operations, operator precedence rules and the linearity of the summation-operator $\sum$. Some of the general identities of the summation operator are often useful. Based upon this information you might be able to provide an explanation of the steps below.

We obtain \begin{align*} \color{blue}{2\sum_{i=0}^{n-1}3^i\left(3^{n-i}-1\right)}&= 2\sum_{i=0}^{n-1}\left(3^n-3^i\right)\\ &=2\cdot3^n\sum_{i=0}^{n-1}1-2\sum_{i=0}^{n-1}3^i\\ &=2\cdot 3^nn-2\cdot\frac{3^n-1}{3-1}\\ &=2\cdot 3^nn-3^n+1\\ &\,\,\color{blue}{=3^n(2n-1)+1} \end{align*}