Question

求连续几个奇数的平方和

1的平方+2的平方+3的平方+-------+n的平方=？

Answer 1

1的平方+2的平方+3的平方+-------+n的平方
=1²+2²+3²+......+n²
=n(n+1)(2n+1)/6
求连续几个奇数的平方和
令S=1²+3²+5²+......(2n-1)²
1²+2²+3²+......+(2n)²
=[1²+3²+5²+......(2n-1)²]+[2²+4²+6²+......+(2n)²]
=[1²+3²+5²+......(2n-1)²]+4[1²+2²+3²+......+n²]
=S+4n(n+1)(2n+1)/6
=S+2n(n+1)(2n+1)/3
由于1²+2²+3²+......+(2n)²
=2n(2n+1)(4n+1)/6
=n(2n+1)(4n+1)/3
所以n(2n+1)(4n+1)/3=S+2n(n+1)(2n+1)/3
S=n(2n+1)(4n+1)/3-2n(n+1)(2n+1)/3
=n[(2n+1)(4n+1)-2(n+1)(2n+1)]/3
=n[(2n+1)(4n+1-2n-2)]/3
=n[(2n+1)(2n-1)]/3
=n(4n²-1)/3

Answer 2

=1²+2²+3²+......+n²
=n(n+1)(2n+1)/6
由于1²+2²+3²+......+(2n)²
=2n(2n+1)(4n+1)/6
=n(2n+1)(4n+1)/3
所以n(2n+1)(4n+1)/3=S+2n(n+1)(2n+1)/3
S=n(2n+1)(4n+1)/3-2n(n+1)(2n+1)/3
=n[(2n+1)(4n+1)-2(n+1)(2n+1)]/3
=n[(2n+1)(4n+1-2n-2)]/3
=n[(2n+1)(2n-1)]/3
=n(4n²-1)/3
懂了吗