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binomial expansion
(x+1)^5=x^5 +5x^4+10x^3+10x^2+5x+1
apply summation from 1 to n
(n+1)^5=5S4+10S3+10S2+5S1+n
we know S3=(n(n+1)/2)^2
and S2=n(n+1)(2n+1)/6 and S1=n(n+1)/2
thus we can find S4
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