求a1=1,an=2a(n-1)+2*(-1)^n (n>=2)的通项公式?

数列{an}满足a1=2,a(n+1)=2an+n+2,求an 数列{an},a1=3,an*a(n+1)=(1/2)^n,求an 已知数列｛an｝中,a(n+1)=an+2^n,a1=3,求an a1=1,a(n+1)=2an+n^2+2n +2 求an 已知a1=2 a(n+1)=2an+2^n+3^n 求an A1=1,A(n+1)/An=(n+2)/n,求An? 1.a1=1 an=a1+2a2+.(n-1)an-1求an2.an+a（n+1）=1/2,a1=1,求an 数列{An}满足a1=1/2,a1+a2+..+an=n方an,求an 已知{an}满足a1=1,a(n+1)=2an+2,求an 已知数列{an},A1=1 A(n+1)=2an/an+2 求a5 数列an满足a1=1,a(n+1)=an/[(2an)+1],求a2010 a(n+1)=2an/3an+4,a1=1/4,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 A1=2 ,A(n+1)=2An+2^n 求An. a(n+1)=an-2*3(n次方)+1,a1=3,求an a1=2,a（n+1）（角标）=an+3n,求an 已知a1=2,an-a(n-1)=n,求an.（看的懂吧>-