# How do you use the summation formulas to rewrite the expression Sigma (4i^2(i-1))/n^4 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?

## sum_(i=1)^n (4i^2(i-1))/n^4 = ((n+1))/(3n^3) ( 3n^2-n-2 ) ##

Let ## S_n = sum_(i=1)^n (4i^2(i-1))/n^4 ## ## :. S_n = 4/n^4sum_(i=1)^n (i^3-i^2) ## ## :. S_n = 4/n^4{sum_(i=1)^n i^3 - sum_(i=1)^n i^2 }##

And using the type results: ## sum_(r=1)^n r^2 = 1/6n(n+1)(2n+1) " ; " sum_(r=1)^n r^3 = 1/4n^2(n+1)^2##

We have;

## S_n = 4/n^4{ 1/4n^2(n+1)^2 - 1/6n(n+1)(2n+1)} ## ## :. S_n = 4/n^4 ( (n(n+1))/12){ 3n(n+1) - 2(2n+1) } ## ## :. S_n = ((n+1))/(3n^3) { 3n^2+3n-4n-2 } ## ## :. S_n = ((n+1))/(3n^3) ( 3n^2-n-2 ) ##

And this has been fitted using Excel for ##n=10, 100, 1000, 10000##

What happens as ##n rarr oo##?

[ NB As an joined function we could haply argue that as ##n rarr oo## then ##S_n rarr 1##; This is probably the misentry of this topic]

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

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

And so,

## lim_(n rarr oo) S_n = lim_(n rarr oo) 1/3( 3+2/n -3/n^2-2/n^3) ## ## :. lim_(n rarr oo) S_n = 1/3( 3+0 -0-0) ## ## :. lim_(n rarr oo) S_n = 1 ##

Which confirms our assumption!