This series

**adam_leeds** · April 1st 2010, 06:33 AM

10 + (10+3) + (10 +2.3) +.....+ (10 +(n-1).3) + (10 +(n+1-1).3)

What can this be simplified to?

**Deadstar** · April 1st 2010, 12:39 PM

Its equal to the following sum.

$\sum_{k=0}^{\infty} 10 + k \cdot 3$

So simplifying this we get...

$\sum_{k=0}^{\infty} 10 + k \cdot 3 = \sum_{k=0}^{\infty} 10 + \sum_{k=0}^{\infty} k \cdot 3$

Taking each part individually...

$\sum_{k=0}^{\infty} 10 = 10n + 10$

$\sum_{k=0}^{\infty} k \cdot 3 = 3\sum_{k=0}^{\infty}k = 3\sum_{k=1}^{\infty}k = 3 \frac{n(n+1)}{2}$

So the total sum is...
$<br /> 10n + 10 + \frac{3n(n-1)}{2} = \frac{1}{2}(3n+20)(n+1)$

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