an=2a(n+1)+6

已知数列{An}中a1=1.且A(n+1)=6n*2^n-An.求通项公试An 在数列{An}中,已知An+A(n+1)=2n (n∈N*) an=2a(n+1)-2 设数列{an}中,a1=2,a(n+1)=an+n+1,求an an=0,an+a(n+1)=2^n,求an通项 已知数列{an}中,a1=1,a(n+1)>an,且[a(n+1)-an]^2-2[a(n+1)+an]+1=0,则an 数列{an},a1=1,a(n+1)=2an-n^2+3n A1=1,A(n+1)/An=(n+2)/n,求An? 已知数列an中,a1=1 2a(n+1)-an=n-2/n(n+1)(n+2) 若bn=an-1/n(n+1) a(n+1)=2an-a(n-1) 3bn-b(n-1)=n An>0,A1=2,当n>=2,An+A(n-1)=n/(An-A(n-1))+2,求An通项 数列an中,a1=6,且an-a(n-1)=a(n-1)/n+n+1,求通项公式