Question

定积分求数列的和

怎么用定积分求数列和,例如：an=x²,结果是sn=﹙1／6﹚n﹙2n+1﹚﹙n+1﹚,具体过程是什么请详细些
数列是特殊的函数，就真不能用定积分求吗

Answer 1

a(n)=n^2=n(n+1)-n=[n(n+1)(n+2)-(n-1)n(n+1)]/3 - [n(n+1)-(n-1)n]/2
s(n) = [1*2*3-0 + 2*3*4-1*2*3 + ...+ (n-1)n(n+1)-(n-2)(n-1)n + n(n+1)(n+2)-(n-1)n(n+1)]/3 - [1*2-0 + 2*3-1*2 + ...+ (n-1)n-(n-2)(n-1) + n(n+1)-(n-1)n]/2
=n(n+1)(n+2)/3 - n(n+1)/2
=(1/6)n(n+1)[2n+4 - 3]
=n(n+1)(2n+1)/6,
定积分可以表达成有限和的极限,可以用来求数列的无穷项的和.