求解递推关系an+2an-1=n+1,n≥1,a0=2

Question

求解递推关系an+2an-1=n+1,n≥1,a0=2

Answer 1

an+2a(n-1)=n+1┈┈┈①,a0=2,a1=-2,a(n+1)+2an=n+2┈┈┈②,②-①得：a(n+1)-an+2[an-a(n-1)]=1,a(n+1)-an-1/3+2[an-a(n-1)-1/3]=0,[a(n+1)-an-1/3]=-2[an-a(n-1)-1/3],[a(n+1)-an-1/3]/[an-a(n-1)-1/3]=-2,数列{a(n+1)-an-1/3}为等比数列,公比q=-2,首项b1=a1-a0-1/3=-2-2-1/3=-13/3,b(n+1)=a(n+1)-an-1/3=-13(-2)^n/3,a(n+1)=n+2-2an,则n+2-2an-an-1/3=-13(-2)^n/3,an=13(-2)^n/9+n/3+5/9.Sn=13[1-(-2)^(n+1)]/27+(3n²+13n+10)/18.