求连续几个奇数的平方和

=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)/3S=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