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I have the following summation: $$\sum_{i=0}^n\left(6in + n(8n+2)\right).$$ Evaluating I get \begin{align*} 6n\left(\frac{n(n+1)}{2}\right) + n(8n+2) = 3n(n(n+1)) + n(8n+2)\\ &=3n^2(n+1) + n(8n+2)\\ &=3n^3 + 3n^2 + 8n^2 + 2n\\ &=3n^3 + 11n^2 + 2n. \end{align*}

Is this correct?

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    Are you starting with $$\left(\sum_{i=0}^{n}6in\right)+n(8n+2)$$ or $$\sum_{i=0}^{n}\left(6in+n(8n+2)\right)?$$2011-03-03
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    the second one.2011-03-03
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    I know that what you want is the value of the sum, but isn't $\sum_{i=0}^n\left(6in + n(8n+2)\right)$ a closed form?2011-03-04

1 Answers 1

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Your first step is wrong.

$$\sum_{i=0}^{n} (6in + n(8n+2)) = \sum_{i=0}^{n} (6in) + \sum_{i=0}^{n} n(8n+2)$$

You wrote

$$\sum_{i=0}^{n} (6in) = 6n \sum_{i=0}^{n} i = 6n^2(n+1)/2$$

but missed this one:

$$\sum_{i=0}^{n} n(8n+2) = n(8n+2) \sum_{i=0}^n 1 = n(n+1)(8n+2)$$

and wrote

$$n(8n+2)$$

instead.

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    in the second summation, are you adding n+1 because the index starts at 0 and not 1?2011-03-03
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    @kry: Yes, exactly! You are adding n+1 times.2011-03-03
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    but when evaluating the first summation, it does not need to be multiplied by (n+1) because you are plugging in for i directly?2011-03-03
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    @Kry: I have edited the answer to clarify it a bit. Since $n$ is independent of $i$, you can take it out of the $\sum$.2011-03-03
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    thanks! that cleared it up2011-03-03