# How do you use the summation formulas to rewrite the expression #Sigma (6k(k-1))/n^3# as k=1 to n without the summation notation and then use the result to find the sum for n=10, 100, 1000, and 10000?

##### 1 Answer

#### Explanation:

Let

# S_n = sum_(k=1)^n 1/n^3 6k(k-1) #

# :. S_n = 6/n^3sum_(k=1)^n (k^2-k) #

# :. S_n = 6/n^3{sum_(k=1)^n (k^2) - sum_(j=1)^n (k)} #

And using the standard results:

We have;

# S_n = 6/n^3{1/6n(n+1)(2n+1) - 1/2n(n+1)} #

# S_n = 1/n^3{n(n+1)(2n+1) - 3n(n+1)} #

# S_n = 1/n^3{n(n+1)(2n+1 - 3)} #

# S_n = 1/n^3 n(n+1)(2n-2) #

# S_n = 2/n^2 (n+1)(n-1) #

# S_n = 2/n^2 (n^2-1) #

And this has been calculated using Excel for

**What happens as #n rarr oo#?**

[ NB As an additional task we could possibly conclude that as

Now,

# :. S_n = 2n^2/n^2 - 2/n^2 #

# :. S_n = 2 - 2/n^2 #

And so,

# lim_(n rarr oo) S_n = lim_(n rarr oo) (2 - 2/n^2) #

# :. lim_(n rarr oo) S_n = lim_(n rarr oo) (2) - lim_(n rarr oo)(2/n^2) #

# :. lim_(n rarr oo) S_n = 2 + 2 lim_(n rarr oo)(1/n^2) #

# :. lim_(n rarr oo) S_n = 2 # ,#" as "(lim_(n rarr oo)(1/n^2) =0)#

Which confirms our assumption!