∑i=1n4i2(i−1)n4=(n+1)3n3(3n2−n−2)∑�=1�4�2(�-1)�4=(�+1)3�3(3�2-�-2)

Let Sn=∑i=1n4i2(i−1)n4��=∑�=1�4�2(�-1)�4 ∴Sn=4n4∑i=1n(i3−i2)∴��=4�4∑�=1�(�3-�2) ∴Sn=4n4{∑i=1ni3−∑i=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)2−16n(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+3n−4n−2}∴��=(�+1)3�3{3�2+3�-4�-2} ∴Sn=(n+1)3n3(3n2−n−2)∴��=(�+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 Sn→1��→1; This is probably the conclusion of this question]

Now, Sn=(n+1)3n3(3n2−n−2)��=(�+1)3�3(3�2-�-2)

∴Sn=13n3(3n3−n2−2n+3n2−n−2)∴��=13�3(3�3-�2-2�+3�2-�-2) ∴Sn=13n3(3n3+2n2−3n−2)∴��=13�3(3�3+2�2-3�-2) ∴Sn=13(3+2n−3n2−2n3)∴��=13(3+2�-3�2-2�3)

And so,

     limn→∞Sn=limn→∞13(3+2n−3n2−2n3)     lim�→∞��=lim�→∞13(3+2�-3�2-2�3) ∴limn→∞Sn=13(3+0−0−0)∴lim�→∞��=13(3+0-0-0) ∴limn→∞Sn=1∴lim�→∞��=1

Which confirms our assumption!