How do you evaluate #\sum _ { n = 1} ^ { 20} ( n + n ^ { 3} ) #?

1 Answer
May 10, 2017

Answer:

# sum_(n=1)^20 (n+n^3) = 44310#

Explanation:

Let:

# S = sum_(n=1)^20 (n+n^3) #

We need the standard formula:

# sum_(r=1)^n r \ = 1/2n(n+1) #
# sum_(r=1)^n r^2 = 1/4n^2(n+1)^2 #

Which gives us:
# S = sum_(n=1)^20 n+sum_(n=1)^20 n^3#
# \ \ = 1/2(20)(21)+1/4(20^2)(21^2)#
# \ \ = 1/2(420)+1/4(400)(441)#
# \ \ = 210 + (100)(441)#
# \ \ = 210 + 44100#
# \ \ = 44310#