An=3An-1+2n+5(A1=1)求An的通项公式用特殊逆推的方法解决

数列｛an},a1=1,an+1=2an-n^2+3n,求｛an｝. an+1 ,n+1是下表 ,an+1=an^3 a1=2 ,求an 数列｛an｝满足a1=1,且an=an-1+3n-2,求an 已知{an}满足a1=1.an=3*n+2An-1,求an 数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an 数列an满足a1=1/3,Sn=n(2n-1)an,求an 数列{an},a1=3,an*a(n+1)=(1/2)^n,求an a1=5,n大于等于2时,an-an-1=6n,求an 已知数列｛an｝中,a(n+1)=an+2^n,a1=3,求an 数列{an}中,a1=1,an+1/an=n/n+2,求an 在｛an｝中,a1=1,an+1=3an+2n,试求｛an｝的通项an a1=1,Sn=n^2an,求an 数列【an】,a1=5,an+1-an=3+4[n-1],求a50 a(n+1)=2an/3an+4,a1=1/4,求an 已知a1=4,an=3an-1+5(n≥),求an a1=2,an+1=2/3an+n-4 求an 数列[An]满足a1=2,a（n+1）=3an-2 求an 在数列{an}中.a1=3且a(n+1)=an^2,求an