2 ^ 2-1 ^ 2=2χ1+1,3 ^ 2-2 ^ 2=2χ2+1,4 ^ 2-3 ^ 2=2χ3+1,…,(n+1) ^ 2-n ^ 2=2χn+1,将以上各式等号两边分别相加得：(n+1) ^ 2-n ^ 2=2χ(1+2+3+…+n)+n
即：1+2+3+…+n=n(n+1)/2。类比上述求法：请你求出1 ^ 2+2 ^ 2+3 ^ 2+…+n ^ 2的公式。
解：令f(n)=1²+2²+3²+……+n²,
a³-b³=(a-b)(a²+ab+b²)，则
2³-1³=2²+1×2+1²,
3³-2³=3²+2×3+2²,
4³-3³=4²+3×4+3²,
……
(n+1)³-n³=(n+1)²+n×(n+1)+n²,
=> (n+1)³-1³=[2²+3²+4²+……+(n+1)²]+[1×2+2×3+3×4+……+n×(n+1)]+f(n);
其中(n+1)³-1³=(n+1-1)[(n+1)²+(n+1)+1²]=n(n²+3n+3);
2²+3²+4²+……+(n+1)²=1²+2²+3²+……+n²-1²+(n+1)²=f(n)+n(n+2);
令g(n)=1×2+2×3+3×4+……+(n-1)×n+n×(n+1),
则g(n)= 1×2+2×3+3×4+……+(n-1)×n+n×(n+1),
错位相加得2g(n)=1×2+2×2²+2×3²+……+2×n²+n×(n+1)=2+n(n+1)+2[f(n)-1²]=2f(n)+n(n+1),
=> g(n)=f(n)+n(n+1)/2;
则得n(n²+3n+3)=[f(n)+n(n+2)]+[f(n)+n(n+1)/2]+f(n)=3f(n)+n(n+2)+n(n+1)/2,
=> f(n)=(2n³+3n²+n)/6,
=> 1²+2²+3²+……+n²=(2n³+3n²+n)/6
a³-b³=(a-b)(a²+ab+b²)，则
2³-1³=2²+1×2+1²,
3³-2³=3²+2×3+2²,
4³-3³=4²+3×4+3²,
……
(n+1)³-n³=(n+1)²+n×(n+1)+n²,
=> (n+1)³-1³=[2²+3²+4²+……+(n+1)²]+[1×2+2×3+3×4+……+n×(n+1)]+f(n);
其中(n+1)³-1³=(n+1-1)[(n+1)²+(n+1)+1²]=n(n²+3n+3);
2²+3²+4²+……+(n+1)²=1²+2²+3²+……+n²-1²+(n+1)²=f(n)+n(n+2);
令g(n)=1×2+2×3+3×4+……+(n-1)×n+n×(n+1),
则g(n)= 1×2+2×3+3×4+……+(n-1)×n+n×(n+1),
错位相加得2g(n)=1×2+2×2²+2×3²+……+2×n²+n×(n+1)=2+n(n+1)+2[f(n)-1²]=2f(n)+n(n+1),
=> g(n)=f(n)+n(n+1)/2;
则得n(n²+3n+3)=[f(n)+n(n+2)]+[f(n)+n(n+1)/2]+f(n)=3f(n)+n(n+2)+n(n+1)/2,
=> f(n)=(2n³+3n²+n)/6,
=> 1²+2²+3²+……+n²=(2n³+3n²+n)/6
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