求公式Formula

1^2 + 2^2 + 3^2 + ... + n^2
= n(n + 1)(2n + 1)/6

1^2 + 3^2 + 5^2 + ... + (2n - 1)^2
= n(2n - 1)(2n + 1)/3

2^2 + 4^2 + 6^2 + ... + 2n^2
= 2n(n + 1)(2n + 1)/3

2013-10-03 19:16:51 補充：
漏咗括號添，第三條係：

2^2 + 4^2 + 6^2 + ... + (2n)^2
= 2n(n + 1)(2n + 1)/3

2013-10-06 07:12:28 補充：
另類方法：利用以下公式
1x2 + 2x3 + 3x4 + ... + n(n+1) = n(n+1)(n+2)/3
所以
1^2 + 3^2 + 5^2 + ... + n^2
= 1x(0+2)/2 + 3x(2+4)/2 + 5x(4+6)/2 + ... + n[(n-1)+(n+1)]/2
= [1x2 + 2x3 + 3x4 + ... + n(n+1)]/2
= [n(n+1)(n+2)/3]/2
= n(n+1)(n+2)/6
而且

2013-10-06 07:15:48 補充：
2^2 + 4^2 + 6^2 + ... + n^2
= 2x(1+3)/2 + 4x(3+5)/2 + 6x(5+7)/2 + ... + n[(n-1)+(n+1)]/2
= [1x2 + 2x3 + 3x4 + ... + n(n+1)]/2
= [n(n+1)(n+2)/3]/2
= n(n+1)(n+2)/6
及
1x2 + 2x3 + 3x4 + ... + (n-1)n = (n-1)n(n+1)/3
所以

2013-10-06 07:20:08 補充：
1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2 + ... + (n-1)^2 + n^2
= n(n+1)(n+2)/6 + (n-1)n(n+1)/6
= n(n+1)(n+2 + n-1)/6
= n(n+1)(2n+1)/6

有 趣！非 常 有 趣！

1^2+2^2+3^2+…+n^2
=n(n+1)(2n+1)/6

1^2+3^2+5^2+...+n^2，(n is odd)
Sol
Set n=2m－1
m=(n+1)/2
1^2+3^2+5^2+...+n^2
=1^2+3^2+5^2+...+(2m－1)^2
=Σ(k=1 to m)_(2k－1)^2
=4Σ(k=1 to m)_k^2－4Σ(k=1 to m)_k+Σ(k=1 to m)_1
=4m(m+1)(2m+1)/6－4m(m+1)/2+m
=m/3*[2(m+1)(2m+1)－6(m+1)+3]
=m*[2(2m^2+3m+1)-6(m+1)+3]/3
=m*(4m^2－1)/3
=[(n+1)/2]*n(n+2)/3
=n(n+1)(n+2)/6

2^2+4^2+6^2+...+n^2，n is even
Sol
n=2m
2^2+4^2+6^2+...+n^2
=2^2+4^2+6^2+...+(2m)^2
=Σ(k=1 to m)_(2k)^2
=4Σ(k=1 to m)_k^2
=4*m(m+1)(2m+1)/6
=2m(2m+2)(2m+1)/6
=n(n+2)(n+1)/6
=n(n+1)(n+2)/6