Creating a formula for $a_n$ for a Fibonacci like sequence

Question

This is sequence is "Fibonacci like":

$$t_1, t_2, t_1+t_2, t_1+2t_2,...$$ How can I find the $1001^{st}$ term of this sequence.

I'm a littler confused because this sequence is neither arithmetic nor geometric so I'm a little confused as to how to solve these problems. I know the Fibonacci sequence and the way to find the numbers for that are:

$f_n=f_{n-1}+f_{n-2}$ and $f_0=1$ and $f_1=1$

How could I approach this problem?

Answer 1

This amounts to solving the recurrence relation

$$ T_1 = t_1, \qquad T_2 = t_2, \qquad T_{n+2} = T_{n+1} + T_n. $$

By linearity, we may write $ T_n = A_n t_1 + B_n t_2 $, where $A_n$ and $B_n$ are sequences defined by the following recurrence relations.

$$ A_1 = 1, \qquad A_2 = 0, \qquad A_{n+2} = A_{n+1} + A_n, \\ B_1 = 0, \qquad B_2 = 1, \qquad B_{n+2} = B_{n+1} + B_n. $$

It is easy to check that $ A_n = F_{n-2}$ and $B_n = F_{n-1} $ solve these, where $F_n$ is the Fibonacci number starting with $F_1 = F_2 = 1$. So

$$ T_n = t_1 F_{n-2} + t_2 F_{n-1}. $$

The followings are first 10 values of $T_n$.

\begin{align*} \begin{array}{|c|c|} \hline n & T_n \\ \hline 1 & t_1 \\ 2 & t_2 \\ 3 & t_1+t_2 \\ 4 & t_1+2 t_2 \\ 5 & 2 t_1+3 t_2 \\ 6 & 3 t_1+5 t_2 \\ 7 & 5 t_1+8 t_2 \\ 8 & 8 t_1+13 t_2 \\ 9 & 13 t_1+21 t_2 \\ 10 & 21 t_1+34 t_2 \\ \hline \end{array} \end{align*}

Answer 2

You look for $r_1, r_2$ roots of $X^2-X-1$, and $\lambda, \mu$ such that $\lambda+\mu=t_1, \lambda.r_1+\mu.r_2=t_2$.

Then you can easily prove that the $n$-th term of the sequence is given by: $$\lambda.r_1^n + \mu.r_2^n$$

Answer 3

From the third term we may find this pattern,

$T_3 = t_1 + t_2 = t_1 + (3-2)t_2$

$T_4 = t_1 + t_3 = t_1 + (4-2)t_2$

If the pattern continues for all consequent terms, then

$$T_n = t_1 + (n-2)t_2$$

So this may be the general $n^{th}$ term.

$T_{1001} = t_1 + 999t_2$