Question

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

Answer 1

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`.

Answer 2

`10120`

Explanation:

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