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n(n+1)(2n+1)/6+(n+1)^2=(n+1)(n+2)(2n+3)/6
We move all terms to the left:
n(n+1)(2n+1)/6+(n+1)^2-((n+1)(n+2)(2n+3)/6)=0
We multiply parentheses ..
n(+2n^2+n+2n+1)/6+(n+1)^2-((n+1)(n+2)(2n+3)/6)=0
We calculate fractions
(n+1)^2=0
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