How do you find the sum of the series #n^2# from n=0 to n=4?

1 Answer
Nov 22, 2016

Answer:

# sum_(n=0)^4 n^2 = 30 #

Explanation:

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

# :. sum_(n=0)^4 n^2 = 1/6(4)(4+1)(8+1) #
# :. sum_(n=0)^4 n^2 = 1/6(4)(5)(9) #
# :. sum_(n=0)^4 n^2 = 30 #

Alternatively, as there are only a few terms we could just write them out and compute the sum;
# sum_(n=0)^4 n^2 = 0^2 + 1^2 +2^2 + 3^2 + 4^2 #
# :. sum_(n=0)^4 n^2 = 1 + 4 + 9 + 16 #
# :. sum_(n=0)^4 n^2 = 30 #, as before