求数列前n项和:1²+3²+5²+……+(2n-1)²=?2²+4²+……+（2n）²=?

题目

求数列前n项和:1²+3²+5²+……+(2n-1)²=?2²+4²+……+（2n）²=?

答案

需要用到平方和公式
1²+2²+3²+.+n²=n(n+1)(2n+1)/6
∵ (2n-1)²=4n²-4n+1
则 1²+3²+5²+……+(2n-1)²
= (4*1²-4+1)+(4*2²-4*2+1)+(4*3²-4*3+1)+.+(4n²-4n+1)
=4*(1²+2²+3²+.+n²)-4(1+2+3+.+n)+n
=4n(n+1)(2n+1)/6-4(n+1)*n/2+n
=2n(n+1)(2n+1)/3-2n(n+1)+n
=(4n³)/3-n/3
∵ (2n)²=4n²
2²+4²+……+（2n）²
=4(1²+2²+3²+.+n²)
=2n(n+1)(2n+1)/3