Problem Solving

Can anyone step by step show me how to simplify this:
4(n(n+1)/2)^2-3n(n+1)/2

Thank you!

sch wrote:Can anyone step by step show me how to simplify this:
4(n(n+1)/2)^2-3n(n+1)/2

Thank you!

if the algebraic notation above is accurate, then a simplification may follow
start from the inner "("
(n+1)
n(n+1)=n^2+n
(n^2+n)/2
[(n^2+n)/2]^2=(n^4+2n^3+n^2)/4
4*(n^4+2n^3+n^2)/4=n^4+2n^3+n^2
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3n(n+1)/2=(3n^2+3n)/2
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n^4+2n^3+n^2 - (3n^2+3n)/2= (2n^4+4n^3+2n^2 - 3n^2-3n)/2= (2n^4+4n^3-n^2-3n)/2

(2n^4+4n^3-n^2-3n)/2= [2(n^2+n)^2 - 2n^2 - n^2 - 3n]/2= [2(n^2+n)^2 - 3(n^2+n)]/2= (n^2+n)(2n^2+2n-3)/2

I'm sorry, I guess we discovered where my algebratic weakness is. Can you explain how you went from 2n^4+4n^3-n^2-3n)/2 to [2(n^2+n)^2 - 2n^2 - n^2 - 3n]/2 ?