0
$\begingroup$

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?

  • 0
    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

3

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.

  • 0
    thanks! that cleared it up2011-03-03