# How do you use the summation formulas to rewrite the expression Σ4i2(i−1)n4Σ4�2(�-1)�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?

i=1n4i2(i1)n4=(n+1)3n3(3n2n2)∑�=1�4�2(�-1)�4=(�+1)3�3(3�2-�-2)

Let Sn=i=1n4i2(i1)n4��=∑�=1�4�2(�-1)�4 Sn=4n4i=1n(i3i2)∴��=4�4∑�=1�(�3-�2) Sn=4n4{i=1ni3i=1ni2}∴��=4�4{∑�=1��3-∑�=1��2}

And using the standard results: r=1nr2=16n(n+1)(2n+1) ; r=1nr3=14n2(n+1)2∑�=1��2=16�(�+1)(2�+1);∑�=1��3=14�2(�+1)2

We have;

Sn=4n4{14n2(n+1)216n(n+1)(2n+1)}     ��=4�4{14�2(�+1)2-16�(�+1)(2�+1)} Sn=4n4(n(n+1)12){3n(n+1)2(2n+1)}∴��=4�4(�(�+1)12){3�(�+1)-2(2�+1)} Sn=(n+1)3n3{3n2+3n4n2}∴��=(�+1)3�3{3�2+3�-4�-2} Sn=(n+1)3n3(3n2n2)∴��=(�+1)3�3(3�2-�-2)

And this has been calculated using Excel for n=10,100,1000,10000�=10,100,1000,10000

What happens as n�→∞?

[ NB As an additional task we could possibly conclude that as n�→∞ then Sn1��→1; This is probably the conclusion of this question]

Now, Sn=(n+1)3n3(3n2n2)��=(�+1)3�3(3�2-�-2)

Sn=13n3(3n3n22n+3n2n2)∴��=13�3(3�3-�2-2�+3�2-�-2) Sn=13n3(3n3+2n23n2)∴��=13�3(3�3+2�2-3�-2) Sn=13(3+2n3n22n3)∴��=13(3+2�-3�2-2�3)

And so,

limnSn=limn13(3+2n3n22n3)     lim�→∞��=lim�→∞13(3+2�-3�2-2�3) limnSn=13(3+000)∴lim�→∞��=13(3+0-0-0) limnSn=1∴lim�→∞��=1

Which confirms our assumption!

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