[tex]\displaystyle\sum_{i=3}^{10}(2i-9)=2\sum_{i=3}^{10}i-9\sum_{i=3}^{10}1[/tex]
[tex]\displaystyle=2\left(\sum_{i=1}^{10}i-\sum_{i=1}^2i\right)-9\left(\sum_{i=1}^{10}-\sum_{i=1}^21\right)[/tex]
Recall that
[tex]\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2[/tex]
[tex]\displaystyle\sum_{i=1}^n1=n[/tex]
so the above reduces to
[tex]2\left(\dfrac{10\cdot11}2-\dfrac{2\cdot3}2\right)-9\left(10-2\right)=32[/tex]
