Find nth term for below sequence

Question

I need nth term for the below sequence

$1,2,(3/2),(5/3),(8/5),(13/8),(21/13),(34/21)........nth term$

I know [(present number)= sum of previous(numerator + denominator ) / previous numerator]

Thanks before

Answer 1

Since the previous denominator is the numerator from two terms before, writing $\frac{N_k}{D_k}$ for the terms of the sequence, we have $N_k=N_{k-1}+N_{k-2}$. This is the Fibonacci sequence relation, and since the first two numerators are $F_2$ and $F_3$ we have $N_k=F_{k+1}$ for every $k$, where $F_k$ is the $k$th Fibonacci number. So the original sequence is $\frac{F_{k+1}}{F_k}$.

Answer 2

Going a little further than @especiallylime, we solve the linear recurrence

$$n_{k+2}=n_{k+1}+n_k,$$ which has the characteristic polynomial

$$t^2-t-1$$ with roots

$$\frac{1\pm\sqrt5}2.$$

We denote the positive root by $\phi$ and the other one is $-\psi=-\phi^{-1}$.

Now the general solution of the recurrence is

$$n_k=c_\phi \phi^k+c_\psi \psi^k,$$ and expressing the initial condition $n_2=1,n_3=2$ (numbering starts at $2$), we obtain

$$n_k=\frac{\phi^k-\psi^k}{\phi-\psi}.$$

The denominators are just one index later, and

$$d_k=\frac{\phi^{k-1}-\psi^{k-1}}{\phi-\psi}.$$

Finally,

$$r_k=\frac{\phi^k-\psi^k}{\phi^{k-1}-\psi^{k-1}}.$$

Answer 3

Let us denote your sequence by $(a_n)_{n \in \mathbb{N}}$. For all $n \in \mathbb{N}$, let's denote by $b_n$ the numerator of $a_n$ and by $c_n$ the denominator of $a_n$.

You can notice that $b_{n+2} = b_{n+1} + b_{n}$ for all $n \in \mathbb{N}$ with $b_0 = 1$ and $b_1 = 2$. This is thus the Fibonacci sequence. The characteristic polynomial of this sequence is $X^2 - X - 1$ and its roots are $$\dfrac{1+\sqrt{5}}{2} \qquad \text{and} \qquad \dfrac{1 - \sqrt{5}}{2}.$$ By using the rules of linear recurrence, you obtain $$b_n = A \left( \dfrac{1+\sqrt{5}}{2} \right)^n + B \left( \dfrac{1-\sqrt{5}}{2} \right)^n$$ for all $n \in \mathbb{N}$, where $A$ and $B$ are constants. Then, using the conditions $b_0 = 1$ and $b_1 = 2$, you find \begin{align*} 1 &= A+B\\ 2 &= A \left(\dfrac{1+\sqrt{5}}{2}\right) + B \left(\dfrac{1-\sqrt{5}}{2}\right) \end{align*} and thus $A=\dfrac{5 + 3\sqrt{5}}{10}, B=\dfrac{5-3\sqrt{5}}{10}$. Finaly, $$b_n = \dfrac{5 + 3\sqrt{5}}{10} \left( \dfrac{1+\sqrt{5}}{2} \right)^n + \dfrac{5-3\sqrt{5}}{10} \left( \dfrac{1-\sqrt{5}}{2} \right)^n$$ for all $n \in \mathbb{N}$.

You can also notice that $c_{n+2} = c_{n+1} + c_n$ for all $n \in \mathbb{N}$, but with the conditions $c_0 = 1, c_1 = 1$. By applying the same technique, you will find a closed formula for $c_n$. Then, finding $a_n$ will be easy.