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?

  • 2
    $\begingroup$ You are here for about $4$ months now and still have not done any effort to use MathJax. This while almost all of your questions were edited by others to make them look better. Let that change please. $\endgroup$
    – drhab
    Dec 17, 2018 at 18:53
  • $\begingroup$ Also note that $\sum_{i=0}^n n = n + \cdots + n = (n+1)n$ and $\sum_{i=0}^n i = 0+1+\cdots+n = \frac{n(\color{red}{n+1})}{2}$. $\endgroup$
    – Christoph
    Dec 17, 2018 at 21:03

1 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*}


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