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I've got the following sum:

$2 \cdot 3^k + \sum\limits_{i=0}^{k-1} 3^{i} \cdot (3 \cdot 2^{k-(i+1)} + 4)$

I know this can be greatly simplified, but I'm not sure how this can be achieved. Any help is greatly appreciated. Thanks.

WolframAlpha simplified it to the following term, but I'm interested in the steps to get there:

$7 \cdot 3^k -3 \cdot 2^k- 2$

Edit:

Thanks for the hint, I think I've figured it out:

$$ \begin{align*} 2 \cdot 3^k + \sum\limits_{i=0}^{k-1} \biggl[3^{i} \cdot (3 \cdot 2^{k-(i+1)} + 4)\biggr] &= 2 \cdot 3^k + \sum\limits_{i=0}^{k-1} \biggl[3^{i+1} \cdot 2^{k-(i+1)} + 4 \cdot 3^i\biggr]\\ &= 2 \cdot 3^k + \sum\limits_{i=0}^{k-1} \biggl[\biggl(\frac{3}{2}\biggr)^{i+1} \cdot 2^{k} + 4 \cdot 3^i\biggr]\\ &= 2 \cdot 3^k + 2^{k} \cdot \sum\limits_{i=0}^{k-1} \biggl[\biggl(\frac{3}{2}\biggr)^{i+1}\biggr] + 4 \cdot \sum\limits_{i=0}^{k-1} \biggl[3^i\biggr]\\ &= 2 \cdot 3^k + \biggl(\frac{3}{2}\biggr) \cdot 2^{k} \cdot \sum\limits_{i=0}^{k-1} \biggl[\biggl(\frac{3}{2}\biggr)^{i}\biggr] + 4 \cdot \sum\limits_{i=0}^{k-1} \biggl[3^i\biggr]\\ &= 2 \cdot 3^k + \biggl(\frac{3}{2}\biggr) \cdot 2^{k} \cdot \frac{1 - \bigl(\frac{3}{2}\bigr)^k}{1-\bigl(\frac{3}{2}\bigr)} + 4 \cdot \frac{1 - 3^k}{1 - 3}\\ &= 2 \cdot 3^k + \frac{\bigl(\frac{3}{2}\bigr) \cdot 2^{k} - 2^{k} \cdot \bigl(\frac{3}{2}\bigr)^{k+1}}{-\bigl(\frac{1}{2}\bigr)} + \frac{4 - 4\cdot3^k}{- 2}\\ &= 4 \cdot 3^k - 3 \cdot 2^{k} + 2^{k+1} \cdot \biggl(\frac{3}{2}\biggr)^{k+1} - 2\\ &= 4 \cdot 3^k - 3 \cdot 2^{k} + 2^{k+1} \cdot \biggl(\frac{3^{k+1}}{2^{k+1}}\biggr) - 2\\ &= 4 \cdot 3^k - 3 \cdot 2^{k} + 3^k \cdot 3 - 2\\ &= 7 \cdot 3^k - 3 \cdot 2^{k} - 2 \end{align*} $$

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  • $\begingroup$ The summation is the sum of three geometric series, Carry out the summations and see what you have. $\endgroup$
    – herb steinberg
    Commented Jun 24, 2021 at 23:53
  • $\begingroup$ You can rework by pulling the constant factors out of the summation $$2\cdot3^k+3\cdot2^{k-1}\sum_{i=0}^{k-1}3^i2^{-i}+4\sum_{i=0}^{k-1}3^i$$ and express the geometric sums. $\endgroup$
    – user65203
    Commented Jun 25, 2021 at 10:20

1 Answer 1

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Congrats for solving the problem, just minor improvement in terms of writing:

  • You might like to write clearer what is the scope of your summation by either using braces or write the summation sign.

    \begin{align*} 2 \cdot 3^k + \sum\limits_{i=0}^{k-1} 3^{i} \cdot (3 \cdot 2^{k-(i+1)} + 4) &= 2 \cdot 3^k + \sum\limits_{i=0}^{k-1} \color{red}(3^{i+1} \cdot 2^{k-(i+1)} + 4 \cdot 3^i \color{red})\\ &= 2 \cdot 3^k + \sum\limits_{i=0}^{k-1} \left[\biggl(\frac{3}{2}\biggr)^{i+1} \cdot 2^{k} + 4 \cdot 3^i \right]\\ \end{align*}

or

\begin{align*} 2 \cdot 3^k + \sum\limits_{i=0}^{k-1} 3^{i} \cdot (3 \cdot 2^{k-(i+1)} + 4) &= 2 \cdot 3^k + \sum\limits_{i=0}^{k-1} 3^{i+1} \cdot 2^{k-(i+1)} + \sum_{i=0}^{k-1}4 \cdot 3^i \\ &= 2 \cdot 3^k + \sum\limits_{i=0}^{k-1} \biggl(\frac{3}{2}\biggr)^{i+1} \cdot 2^{k} + \sum\limits_{i=0}^{k-1}4 \cdot 3^i \\ \end{align*}

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  • $\begingroup$ Thanks! I'm quite happy that I've figured it out. Thanks for your advice! I've now improved my equation for better readability. $\endgroup$
    – iHell
    Commented Jun 25, 2021 at 10:10

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