How do you find the partial sum of #Sigma (250-8/3i)# from i=1 to 60?

2 Answers
Jan 4, 2018

Answer:

The answer is 10120 (see below).

Explanation:

First, the commutative and distributive properties allow us to write:

#sum_{i=1}^{60}(250-8/3 i)=sum_{i=1}^{60}250-8/3 sum_{i=1}^{60}i#.

Now #sum_{i=1}^{60}250# is just 250 added to itself 60 times. Therefore #sum_{i=1}^{60}250=60*250=15000#.

Next, we can use the well-known formula for the sum of the first #n# integers (see https://brilliant.org/wiki/sum-of-n-n2-or-n3/), which is #sum_{i=1}^{n}i=1+2+3+cdots+n=(n(n+1))/2#, to say

#sum_{i=1}^{60}i=(60*61)/2=30*61=1830#.

Hence, the answer is

#sum_{i=1}^{60}250-8/3 sum_{i=1}^{60}i#

#=15000-8/3 * 1830=15000-4880=10120#.

Jan 4, 2018

Answer:

#10120#

Explanation:

#sum_{i=1}^60 (250-8/3*i)=250*60-8/3*(60*61)/2=10120#