TO DERIVE :
let,
S=i=1∑ni3 ............(1)
We start off the proof with the "telescoping sum" (or the collapsing sum).
i=1∑n[(i+1)4−i4] SOLVING IN GENERAL FORM
=(24−14)+(34−24)+(44−34)+...........+................+(n+1)4−n4+...........+................+(n)4−(n−1)4=(n+1)4−14=(n=1)4−1=((n+1)2)2−1=(n2+2n+1)2−1
i=1∑n[(i+1)4−i4] =n4+4n3+6n2+4n+1 .......................(2)
now,
i=1∑n[(i+1)4−i4] SOLVING IN terms of i
=i=1∑n[(i4+4i3+6i2+4i+1)−i4]=i=1∑n[(4i3+6i2+4i+1)]=4i=1∑ni3+6i=1∑ni2+4i=1∑ni+i=1∑n1
from (1) ..further solving above...
=4S+6(6n(n+1)(2n+1))+4(2n(n+1))+n
=4S+2n3+n2+2n2+n+2n2+3n=4S+2n3+5n2+4n ................(3)
THUS; equating (2) and (3);
n4+4n3+6n2+4n=4S+2n3+5n2+4nn4+2n3+n2=4Sn2(n2+2n+1)=4S
4S=n2(n+1)2S=22n2(n+1)2
S=[2n(n+1)]2
hence drived.
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