Show that $z_{k-1}z_{k+1}=z_k+1 \rightarrow z_5=\frac{z_1+1}{z_2}$

Question

So given the recurrence relation:$$z_{k-1}z_{k+1}=z_k+1$$

I have that $z_3=\frac{z_2+1}{z_1}$ and $z_4=\frac{z_1+z_2+1}{z_1z_2}$, And apparently the notes I'm following it says that $z_5=\frac{z_1+1}{z_2}$ but I can't get the algebra to work out.. Can somebody help me out? Thanks

Answer 1

By the recurrence relation, you get: $$ z_5=\frac{z_4 + 1}{z_3} $$ Note that: $$ z_4 + 1= \frac{z_1 + z_2 + z_1 z_2 + 1}{z_1 z_2} = \frac{(z_1 + 1) + z_2(z_1 +1)}{z_1 z_2} = \frac{(z_1 +1)(z_2 + 1)}{z_1 z_2} $$ Then: $$ z_5 = \frac{(z_1 +1)(z_2 + 1)}{z_1 z_2} \cdot \frac{z_1}{1 + z_2} = \frac{z_1 + 1}{z_2} $$

Answer 2

$z_3 = \frac{z_2+1}{z_1}$ and $z_4 = \frac{z_3 + 1 }{z_2} = \frac{z_2 + 1 + z_1}{z_1z_2}$

So:

$z_5 = \frac{z_4+1}{z_3} = z_1 \frac{z_4+1}{z_2+1} = z_1 (z_4+1) \frac{1}{z_2+1} = z_1 (\frac{z_2+1+z_1}{z_1z_2}+1)\frac{1}{z_2+1} = z_1 \frac{z_2+z_1+1+z_1z_2}{z_1z_2} \frac{1}{1+z_2} = \frac{(z_2+1)(z_1+1)}{z_2(1+z_2)}$

which gives $z_5 = \frac{z_1+1}{z_2}$

Answer 3

$$z_k=\frac{z_{k-1}+1}{z_{k-2}}$$ $$z_5=\frac{z_4+1}{z_3}=\left(\frac{z_3+1}{z_2}+1\right)/\left(\frac{z_2+1}{z_1}\right)= z_1(z_3+z_2+1)/z_2(z_2+1)$$ $$=z_1(\frac{z_2+1}{z_1}+z_2+1)/z_2(z_2+1)=\frac{z_1\left(\frac{1}{z_1}+1\right)}{z_2}=\frac{z_1+1}{z_2}$$