00:01
In this problem, we want to evaluate the following sum.
00:05
The sum from k is equal to 1 to n of 6k times k minus 1 divided by n to 3.
00:16
To violate this limit, we need to recall of two smaller identities.
00:24
The sum of i is equal to 1 to i of i is equal to n times n plus 1 divided by 2.
00:42
The next sum that we're going to require the sum for i is equal to 1 to n of i square and this is equal to n times n plus 1 times 2n plus 1 divided by 6.
01:07
So we're going to use these two known results to solve our problem.
01:16
So let's re -express our sum as n as two smaller sums.
01:25
But first we can factor out the 6 and the n -cube as is not going to affect our sum over k.
01:33
So we will have that our sum is equal to 6 over n to the power of 3 times the sum from k to n of k squared minus k.
01:52
I've simply factored in the k squared.
01:55
So now this separates into two smaller sums.
02:08
Sum of k squared minus the sum of k.
02:16
And we have these known results.
02:19
The only changes that our variable is different.
02:23
So this is going to give us 6 divided by n -cube of n times n plus 1 times 2 n plus 1 divided by 6 minus n times n plus 1.
02:54
Divided by two.
02:58
Okay, let's evaluate these fractions.
03:03
The first thing we want to do is factor out the n plus 1, and then n.
03:07
So we will have.....