n=1/(xn-2)+1/3=(xn+1)/3(xn-2)
b(n+1)=1/[x(n+1)-2]+1/3
=1/[f(xn)-2]+1/3
=1/[(1+2/xn)-2]+1/3
=2(xn+1)/3(2-xn)
b(n+1)/bn=-2(定值)
b1=1/(11/7-2)+1/3=-2
所以bn是以-2为首项,公比为-2的等比数列
n=1/(xn-2)+1/3=(xn+1)/3(xn-2)
b(n+1)=1/[x(n+1)-2]+1/3
=1/[f(xn)-2]+1/3
=1/[(1+2/xn)-2]+1/3
=2(xn+1)/3(2-xn)
b(n+1)/bn=-2(定值)
b1=1/(11/7-2)+1/3=-2
所以bn是以-2为首项,公比为-2的等比数列