数列{an}满足a1=2,na(n+1)-3(n+1)an=-2n,则an=
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数列{an}满足a1=2,na(n+1)-3(n+1)an=-2n,则an=
数列{an}满足a1=2,na(n+1)-3(n+1)an=-2n²-4n-3,则an=
不好意思
数列{an}满足a1=2,na(n+1)-3(n+1)an=-2n²-4n-3,则an=
不好意思
na(n+1)-3(n+1)an=-2n²-4n-3
na(n+1)-n²-2n
=3(n+1)an-3n²-6n-3
=3(n+1)an-3(n+1)²
=3(n+1)(an-n-1)
即n[a(n+1)-(n+1)-1]=3(n+1)(an-n-1)
[a(n+1)-(n+1)-1]/(n+1)=3(an-n-1)/n
设数列{bn},令bn=(an-n-1)/n
则有b(n+1)=3bn
所以{bn}是等比数列
又b1=a1-1-1=0
所以bn=b1*q^(n-1)=0
即(an-n-1)/n=0
所以an=n+1
na(n+1)-n²-2n
=3(n+1)an-3n²-6n-3
=3(n+1)an-3(n+1)²
=3(n+1)(an-n-1)
即n[a(n+1)-(n+1)-1]=3(n+1)(an-n-1)
[a(n+1)-(n+1)-1]/(n+1)=3(an-n-1)/n
设数列{bn},令bn=(an-n-1)/n
则有b(n+1)=3bn
所以{bn}是等比数列
又b1=a1-1-1=0
所以bn=b1*q^(n-1)=0
即(an-n-1)/n=0
所以an=n+1
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