Question

求和Sn=1²－2²＋3²－4²……﹙﹣1﹚的n－2次方×n²

Answer 1

当n为偶数时：

1²－2²＋3²－4²……﹙﹣1﹚的n+1次方×n²

= (1+2)(1-2) + (3+4)(3-4) + (5+6)(5-6) + .+ (n-1+n)(n-1-n)

= -(1+2+3+.+n)

= -n(n+1)/2

当n为奇数时：

1²－2²＋3²－4²……﹙﹣1﹚的n+1次方×n²

= 1 - (2+3)(2-3) - (4+5)(4-5) - ...- (n-1+n)(n-1-n)

= 1+2+3+.+n

= n(n+1)/2