Question

Find the sum of this problem

Options:
52

 
26  
13  
32

Answer 1

`\displaystyle\sum_(i=3)^(10)(2i-9)=2\sum_(i=3)^(10)i-9\sum_(i=3)^(10)1`

`\displaystyle=2\left(\sum_(i=1)^(10)i-\sum_(i=1)^2i\right)-9\left(\sum_(i=1)^(10)-\sum_(i=1)^21\right)`

Recall that


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

`\displaystyle\sum_(i=1)^n1=n`

so the above reduces to


`2\left(\frac{10\cdot11}2-\frac{2\cdot3}2\right)-9\left(10-2\right)=32`