How do you find the partial sum of #Sigma (1000-5n)# from n=0 to 50?

1 Answer
Feb 18, 2017

Answer:

# sum_(n=0)^50 (1000-5n) = 44625 #

Explanation:

# sum_(n=0)^50 (1000-5n) = sum_(n=0)^50 (1000) - 5sum_(n=0)^50(n) =#

We can use the standard result #sum_(i=1)^n i = 1/2n(n+1) #, and note that:

#sum_(n=0)^50 (1000) = 1000+1000+ ... +1000 \ \# (51 terms)
#sum_(i=0)^n i=0+ sum_(i=1)^n #

So we get:

# sum_(n=0)^50 (1000-5n) = (51)(1000) - 5*1/2(50)(50+1)#
# " "= 51000 - 5/2(50)(51)#
# " "= 51000 - 6375#
# " " = 44625#