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?

1 Answer
Dec 15, 2016

Answer:

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

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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#

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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!