How do you find the partial sum of #Sigma (1000-n)# from n=1 to 250?

1 Answer
Mar 26, 2018

Answer:

# sum_(r=1)^250 (1000-r) = 218625 #

Explanation:

Let us denote the sum:

# S_n = sum_(r=1)^n (1000-r) #

Then we can use the standard result for the sum of the first #n# integers (Using the formula for Arithmetic progression):

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

And so we get:

# S_n = sum_(r=1)^n (1000) - sum_(r=1)^n(r) #
# \ \ \ \ = 1000n - 1/2n(n+1) #

# \ \ \ \ = n/2{2000 - (n+1)} #

# \ \ \ \ = n/2(2000-n-1) #

# \ \ \ \ = n/2(1999-n) #

Using this result, the desired sum is:

# sum_(r=1)^250 (1000-r) = S_250#
# " " = 250/2(1999-250) #
# " " = 125(1749) #
# " " = 218625 #