證明1^2+2^2+3^2+...+n^2=n(n+1)(2n+1)/6 不用數學歸立法!?

Sol
P=1^2+2^2+3^2+...+n^2=Σ(k=1 to n)_k^2
A=2^3+3^3+4^3+…+(n+1)^3=Σ(k=1 to n)_(k+1)^3
B=1^3+2^3+3^3+…+n^3=Σ(k=1 to n)_k^3
從左
A-B=(n+1)^3-1=n^3+3n^2+3n
從右
A-B=Σ(k=1 to n)_(k+1)^3-Σ(k=1 to n)_k^3
=Σ(k=1 to n)_(k+1)^3-Σ(k=1 to n)_k^3
=Σ(k=1 to n)_[(k+1)^3-k^3]
=Σ(k=1 to n)_(3k^2+3k+1)
=Σ(k=1 to n)_3k^2+Σ(k=1 to n)_3k+Σ(k=1 to n)_1
=3P+3n(n+1)/2+n
左=右n^3+3n^2+3n=3P+3n(n+1)/2+n
2n^3+6n^2+6n=6P+3n(n+1)+2n
6P=2n^3+6n^2+6n－3n^2－3n－2n
=2n^3+3n^2+n
=n(2n^2+3n+1)
=n(n+1)(2n+1)
P=n(n+1)(2n+1)/6