Answer:
[tex]\sum^{n=9}_{n=2} (-3+5n)=196[/tex]
Step-by-step 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,
[tex]y=\sum^{n=9}_{n=2} (-3+5n)[/tex]
Solve,
[tex]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))[/tex]
[tex]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))[/tex]
[tex]y=(-3+10)+(-3+15)+(-3+20)+(-3+25)+(-3+30)+(-3+35)+(-3+40)+(-3+45)[/tex]
[tex]y=7+12+17+22+27+32+37+42[/tex]
[tex]y=196[/tex]
Therefore, [tex]\sum^{n=9}_{n=2} (-3+5n)=196[/tex]
