题目

己知数列｛an｝满足a1=3,an+1=an+3n2+3n+2一1/n（n+1）,n£N

求数列an的通项公式


证明 1/a1+1/a2+L+1/an＜1/2






(1)
a(n+1)=an+3n²+3n+2 -1/[n(n+1)]=an+(n+1)³-n³-1/n +1/(n+1) +1
[a(n+1) -(n+1)³ -1/(n+1)]-(an-n³ -1/n)=1,为定值
a1 -1³ -1/1=3-1-1=1
数列{an -n³ -1/n}是以1为首项,1为公差的等差数列
an -n³ -1/n=1+1×(n-1)=n
an=n³+n +1/n
n=1时,a1=1+1 +1/1=3,同样满足通项公式
数列{an}的通项公式为an=n³+n +1/n
(2)
an=n³+n +1/n=(n⁴+n²+1)/n
 1/an= n/(n⁴+n²+1)  算到这步我就不会求和了  怎么办

提问时间：2021-04-02

答案

a(n+1) = a(n) + 3n^2 + 3n + 2 - 1/[n(n+1)]= a(n) + 3n(n+1) + 2(n+1-n) - 1/n + 1/(n+1),= a(n) + [n(n+1)(n+2) - (n-1)n(n+1)] + 2(n+1) - 2n - 1/n + 1/(n+1),a(n+1) - n(n+1)(n+2) - 2(n+1) - 1/(n+1) = a(n) ...