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