# 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?

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:

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

What happens as

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

Now,

# :. 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!

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To rewrite the expression Σ (4i^2(i-1))/n^4 as k=1 to n without summation notation, we first expand and simplify the expression:

Σ (4i^2(i-1))/n^4 = 4/n^4 * Σ (i^3 - i^2) from i=1 to n

Using the summation formulas, we get:

Σ (i^3 - i^2) = (Σ i^3) - (Σ i^2)

Now, apply the summation formulas for the sum of the first n integers (Σ i) and the sum of the squares of the first n integers (Σ i^2):

Σ i = (n(n+1))/2 Σ i^2 = (n(n+1)(2n+1))/6 Σ i^3 = [(n(n+1))/2]^2

Substitute these formulas into the expression:

Σ (4i^2(i-1))/n^4 = 4/n^4 * [(n(n+1))/2]^2 - (n(n+1)(2n+1))/6

Now, plug in the values of n to find the sum for n=10, 100, 1000, and 10000:

For n=10: Substitute n=10 into the expression and calculate the result. For n=100: Substitute n=100 into the expression and calculate the result. For n=1000: Substitute n=1000 into the expression and calculate the result. For n=10000: Substitute n=10000 into the expression and calculate the result.

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