How do you evaluate #\sum _ { k = 0} ^ { n } ( 3- 2k )#?
1 Answer
Jun 25, 2017
# sum_(k = 0)^n ( 3- 2k ) = (3-n)(n+1)#
Explanation:
Using the standard summation formula:
# sum_(k = 1)^n k = 1/2n(n+1)#
We have:
# sum_(k = 0)^n ( 3- 2k ) = sum_(k = 0)^n 3-2sum_(k = 0)^n k #
# " " = sum_(k = 0)^n 3-2sum_(k = 1)^n k #
# " " = 3(n+1) - 2 \ 1/2n(n+1)#
# " " = 3(n+1) - n(n+1)#
# " " = (3-n)(n+1)#
