数列{bn}满足 3bn+1 + 3bn-1 = bn,b1 =1,求{bn}的通项公式
问题描述:
数列{bn}满足 3bn+1 + 3bn-1 = bn,b1 =1,求{bn}的通项公式
答
3b(n+1) -bn+ 3b(n-1) = 0设3[b(n+1)-xbn] -y[bn-xb(n-1)] = 03b(n+1)-(3x+y)bn+xyb(n-1) = 03x+y=1,xy=3x1=(1+i√11)/2,y1=-(1+i3√11)/2x2=(1-i√11)/2,y2=-(1-i3√11)/23[b(n+1)-x1bn]=y1[bn-x1b(n-1)]3[b(n+1)-x...