How do you evaluate the series #Sigma 5r# from r=3 to 8?

1 Answer
Jan 3, 2017

Answer:

#sum_(r=3)^8 5r = 165#

Explanation:

We seek:

# sum_(r=3)^8 5r = 5sum_(r=3)^8 r #

Due to the small number of terms required we can just expand the individual terms to get:

# sum_(r=3)^8 5r = 5(3+4+5+6+7+8} #
# " " = 5(33) #
# " " = 165 #

If the number of individual terms were larger this would be quite cumbersome and use of the standard summation formula

# sum_(r=1)^n r = 1/2n(n+1) #

would be more appropriate. If we used this approach we would get

# sum_(r=3)^8 5r = 5{(sum_(r=1)^8) r - (sum_(r=1)^2 r )}#
# " " = 5{1/2*8*9 - 1/2*2*3} #
# " " = 5/2{72 - 6} #
# " " = 5/2*66 #
# " " = 165 #