How do you find the sum of the series #k^2-1# from k=1 to 5?

1 Answer
Nov 10, 2016

Answer:

# sum_(k=1)^5 (k^2 - 1) = 50 #

Explanation:

We need the standard formula #sum_(r=1)^n r^2=1/6n(n+1)(2n+1)#

# :. sum_(k=1)^5 (k^2 - 1) = sum_(k=1)^5 (k^2) - sum_(k=1)^5 (1) #
# :. sum_(k=1)^5 (k^2 - 1) = 1/6(5)(5+1)(10+1) - 5 #
# :. sum_(k=1)^5 (k^2 - 1) = 1/6(5)(6)(11) - 5 #
# :. sum_(k=1)^5 (k^2 - 1) = 55 - 5 #
# :. sum_(k=1)^5 (k^2 - 1) = 50 #

Alternatively, as there are only a few terms we could just write them out and compute the sum;
# sum_(k=1)^5 (k^2 - 1) = (1-1) + (4-1) + (9-1) + (16-1) + (25-1) #
# :. sum_(k=1)^5 (k^2 - 1) = 0 + 3 + 8 + 15 + 24 #
# :. sum_(k=1)^5 (k^2 - 1) = 50 #, as before