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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