Question

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?

Answer 1

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

Explanation:

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 standard 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 calculated using Excel for `n=10, 100, 1000, 10000`

What happens as `n rarr oo`?

[ NB As an additional task we could possibly conclude that as `n rarr oo` then `S_n rarr 1`; This is probably the conclusion of this question]

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!