Question

(Calculus help, 25 points, will give brainliest) Find the sum of the sigma notation below.

Answer 1

$\displaystyle\sum_{i=3}^{10}(2i-9)=2\sum_{i=3}^{10}i-9\sum_{i=3}^{10}1$

Recall that

$\displaystyle\sum_{i=1}^n1=n$

$\displaystyle\sum_{i=1}^ni=\frac{n(n+1)}2$

Your sums start at $i=3$, so in order to apply these formulas directly, you need to compensate for the missing first two terms:

$\displaystyle\sum_{i=3}^{10}1=\sum_{i=1}^{10}1-(1+1)=10-2=8$

$\displaystyle\sum_{i=3}^{10}i=\sum_{i=1}^{10}i-(1+2)=\frac{10\cdot11}2-3=52$

$\implies\displaystyle\sum_{i=3}^{10}(2i-9)=2\cdot52-9\cdot8=\boxed{32}$