Precalculus, Chapter 9, 9.4, Section 9.4, Problem 54

You need to evaluate the given sum using formulas for the sums of powers of integers, such that:
sigma_(j=1)^10 (3 - (1/2)j +(1/2)j^2) = sigma_(j=1)^10 3 - (1/2)sigma_(j=1)^10 j + (1/2)sigma_(j=1)^10 j^2
sigma_(j=1)^10 3 = 3 + 3 + .. + 3 = 10*3 = 30
sigma_(j=1)^10 j = 1 + 2 + .. + 10
You need to remember that sigma_(n=1)^n k = ((n(n+1))/2)
Replacing 10 for j yields:
sigma_(j=1)^10 j = 10(10+1)/2 = 5*11
sigma_(j=1)^10 j^2 = 1^2 + 2^2 + ...+ 10^2
You need to remember that sigma_(n=1)^n k^2 = ((n(n+1)(2n+1))/6)
Replacing 10 for n yields:
sigma_(j=1)^10 j^2 = ((10(10+1)(2*10+1))/6)
sigma_(j=1)^10 j^2 = (10*11*21)/6 = 5*11*7 = 35*11
Evaluating the sum, yields:
sigma_(j=1)^10 (3 - (1/2)j +(1/2)j^2) = 30 - (1/2)*5*11 + (1/2)*35*11
sigma_(j=1)^10 (3 - (1/2)j +(1/2)j^2) = (60 - 55 + 385)/2 = 195
Hence, evaluating the given sum using formulas for the sums of powers of integers, yields sigma_(j=1)^10 (3 - (1/2)j +(1/2)j^2) = 195.