Continuous sum of squares, such as 1 + 2 + 3 + +N, what's the final formula,

Continuous sum of squares, such as 1 + 2 + 3 + +N, what's the final formula,

1^2=1/6*1(2*1+1)(1+1)=1/6*6=1
1^2+2^2=1/6*(2*2+1)(2+1)=1/6*30=5
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Suppose 1 + 2 + 3 + +N = 1 / 6N (2n + 1) (n + 1)
be
1^2+2^2+3^2+…… +n^2+(n+1)^2
=1/6n(2n+1)(n+1)+(n+1)^2
=1/6(n+1)(2n^2+n+6n+6)
=1/6*(n+1)(2n+3)(n+2)
=1/6*(n+1)[2(n+1)+1][(n+1)+1]
The hypothesis holds
Get proof