What is ∑n=132(4n+1) equal to?

n = 1  --->     4(1) + 1  =  5

n = 2  --->     4(2) + 1  =  9

n = 3  --->     4(3) + 1  =  13

...

n = 32 -->     4(32) + 1  =  128

This is an arithmetic series with:

a = first term = 5

d = common difference = 4

l = last term = 128

n = number of terms = 32

Sum of an arithmetic series:  Sum  =  n(a + l) / 2

Place the numbers into the equation and solve.

Geno3141: A slight typo: Last term =128+1=129