Question

Solve the summation from n equals 2 to 9 of negative 3 plus 5 times n.

Answer 1

`\sum^(n=9)_(n=2) (-3+5n)=196`

Explanation:

Given : Expression the summation from n equals 2 to 9 of negative 3 plus 5 times n.

To find : Solve the expression ?

Solution :

The summation from n equals 2 to 9 of negative 3 plus 5 times n is written as,

`y=\sum^(n=9)_(n=2) (-3+5n)`

Solve,

`y=(-3+5(2))+(-3+5(3))+(-3+5(4))+(-3+5(5))+(-3+5(6))+(-3+5(7))+(-3+5(8))+(-3+5(9))`

`y=(-3+5(2))+(-3+5(3))+(-3+5(4))+(-3+5(5))+(-3+5(6))+(-3+5(7))+(-3+5(8))+(-3+5(9))`

`y=(-3+10)+(-3+15)+(-3+20)+(-3+25)+(-3+30)+(-3+35)+(-3+40)+(-3+45)`

`y=7+12+17+22+27+32+37+42`

`y=196`

Therefore,
`\sum^(n=9)_(n=2) (-3+5n)=196`