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having a bit of trouble working through sigma notation in first year calc. I get the basic principles at work here, but when the stopping point is a variable, like n, I don't really know how to simplify. Couldn't find answers elsewhere because I didn't quite know how to articulate my question!

So this is the question I'm working with, says to find the value of the sum.

$$\sum_{i=1}^n (3 + 6i)^2$$

How would I evaluate this and then notate my answer? Cheers!

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    $\begingroup$ Do you know formulas for a) $1+2+\cdots+n$, and b) $1^2+2^2+\cdots+n^2$? $\endgroup$ – angryavian Nov 29 '17 at 7:28
  • $\begingroup$ Yep, I know that much. It's really more a question of applying them together here? $\endgroup$ – Avi Nov 29 '17 at 7:29
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The final answer is in terms of $n$.

\begin{align} \sum_{i=1}^n (3+6i)^2 &=\sum_{i=1}^n(3^2+2(3)(6i)+(6i)^2) \\ &=\sum_{i=1}^n(9+36i+36i^2)\\ &=\sum_{i=1}^n 9 + 36\sum_{i=1}^n i + 36 \sum_{i=1}^ni^2 \end{align}

I hope you can evaluate the last expression.

Alternative:

\begin{align}\sum_{i=1}^n (3+6i)^2&=9\sum_{i=1}^n(1+2i)^2 \\ &= 9 \left( -1+\sum_{i=0}^n(1+2i)^2\right)\\ &=9\left( -1 + \sum_{i=1}^{2n+1} i^2 - \sum_{i=1}^n(2i)^2\right)\end{align}

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