递推与数列问题设数列{an}的前n项和为Sn,若a1=1/2,Sn=n^2*an-n(n-1),试写出Sn与Sn-1(n>=2)的_数学解答

递推与数列问题
设数列{an}的前n项和为Sn,若a1=1/2,Sn=n^2*an-n(n-1),试写出Sn与Sn-1(n>=2)的递推关系式,并求出Sn关于n的表达式.

人气：405 ℃　时间：2020-05-22 12:21:10

解答

Sn=n^2*an-n(n-1)=n^2*(Sn-S(n-1))-n(n-1)(n^2-1)Sn-n^2*S(n-1)-n(n-1)=0(n+1)Sn-n^2*S(n-1)/(n-1)-n=0((n+1)/n)Sn=(n/(n-1))S(n-1)+1Sn=(n^2/(n+1)(n-1))S(n-1)+n/(n+1)2S1=2*a1=2*(1/2)=1((n+1)/n)Sn=nSn=n^2/(n+1...