10 + (10+3) + (10 +2.3) +.....+ (10 +(n-1).3) + (10 +(n+1-1).3)
What can this be simplified to?
Its equal to the following sum.
$\displaystyle \sum_{k=0}^{\infty} 10 + k \cdot 3$
So simplifying this we get...
$\displaystyle \sum_{k=0}^{\infty} 10 + k \cdot 3 = \sum_{k=0}^{\infty} 10 + \sum_{k=0}^{\infty} k \cdot 3$
Taking each part individually...
$\displaystyle \sum_{k=0}^{\infty} 10 = 10n + 10$
$\displaystyle \sum_{k=0}^{\infty} k \cdot 3 = 3\sum_{k=0}^{\infty}k = 3\sum_{k=1}^{\infty}k = 3 \frac{n(n+1)}{2}$
So the total sum is...
$\displaystyle
10n + 10 + \frac{3n(n-1)}{2} = \frac{1}{2}(3n+20)(n+1)$