The summation expression in the following series has an absolute value in it. Expand and evaluate the summation notation. What is the sum of the series?

[tex]|3i-10|=\begin{cases}3i-10&\text{for }3i-10\ge0\\-(3i-10)&\text{for }3i-10<0\end{cases}=\begin{cases}3i-10&\text{for }i\ge\dfrac{10}3\\\\10-3i&\text{for }i<\dfrac{10}3\end{cases}[/tex]

We have that [tex]3<\dfrac{10}3<4[/tex], which means [tex]|3i-10|[/tex] reduces to [tex]3i-10[/tex] when [tex]i\ge4[/tex], or reduces to [tex]10-3i[/tex] when [tex]i\le3[/tex].

So we can expand the summation as

[tex]\displaystyle\sum_{i=1}^6|3i-10|=\sum_{i=1}^3(10-3i)+\sum_{i=4}^6(3i-10)[/tex]

Notice that the 10 contributes a total of 30 from the first sum, and -30 from the second sum, so those terms cancel, leaving us with

[tex]\displaystyle3\left(\sum_{i=4}^6i-\sum_{i=1}^3i\right)=3((4+5+6)-(1+2+3))=3(15-6)=3(9)=27[/tex]