Both Fibonacci and Lucas sequence are described by

$t_n=t_{n-1}+t_{n-2}$

But they differ only in their first 2 terms. Here, we will make our own sequence by changing the first two terms.

We will choose $t_1=2$, $t_2=3$

$t_3=2+3=5$

$t_4=3+5=8$

$t_5=5+8=13$

$t_6=8+13=21$

$t_7=13+21=34$

$t_8=21+34=55$

Now, we shall find observe the ratios of the consecutive terms.

$dfrac{t_4}{t_3}=dfrac{8}{5}=1.6$

$dfrac{t_5}{t_4}=dfrac{13}{8}=1.625$

$dfrac{t_6}{t_5}=dfrac{21}{13}=1.6154$

$dfrac{t_7}{t_6}=dfrac{34}{21}=1.6159$

$dfrac{t_8}{t_7}=dfrac{55}{34}=1.6176$

Notice that ratios tend to approach $1.618approx dfrac{1+sqrt{5}}{2}$

This is the same ratio as that of Fibonacci and Lucas sequences.

The ratios of consecutive terms tends to approach $dfrac{1+sqrt{5}}{2}$ which is the same value as that of Fibonacci and Lucas sequences.

The recursive formula for Fibonacci sequence
$t_n=t_{n-1}+t_{n-2}$

Notice that for a geometric sequence with common ratio $r$, the terms are

$a,;ar^2,;ar^3,;ar^4$

Combining with the recursive formula for Fibonacci sequence

$t_n=t_{n-1}+t_{n-2}$

We shall have

$ar^n=ar^{n-1}+ar^{n-2}$

For $n=2$

$r^2=r+1$

$r^2-r+1=0$

This is a quadratic equation with $a=1$, $b=-1$, and $c=1$

which we shall solve using quadratic formula

$r=dfrac{-bpmsqrt{b^2-4ac}}{2a}$

$r=dfrac{1pmsqrt{(-1)^2-4(1)(1)}}{2(1)}=dfrac{-1pmsqrt{5}}{2}$

The value of the positive root, is $approx 1.618$ which is the common ratios of consecutive terms for Fibonacci and Lucas sequences.

$$
1.618
$$

a) We will obtain the first 10 terms using the given recursive formula.

$t_1=1$

$t_2=5$

$t_3=5+2(1)=7$

$t_4=7+2(5)=17$

$t_5=17+2(7)=31$

$t_6=31+2(17)=65$

$t_7=65+2(31)=127$

$t_8=127+2(65)=257$

$t_9=257+2(127)=511$

$t_10=511+2(257)=1025$

b) The ratios of the consecutive terms are as follows.

$dfrac{t_2}{t_1}=5$

$dfrac{t_3}{t_2}=dfrac{7}{5}=1.4$

$dfrac{t_4}{t_3}=dfrac{17}{7}=2.4286$

$dfrac{t_5}{t_4}=dfrac{31}{17}=1.8235$

$dfrac{t_6}{t_5}=dfrac{65}{31}=2.0968$

$dfrac{t_7}{t_6}=dfrac{127}{65}=1.9538$

$dfrac{t_8}{t_7}=dfrac{257}{127}=2.0236$

$dfrac{t_9}{t_8}=dfrac{511}{257}=1.9883$

$dfrac{t_{10}}{t_9}=dfrac{1025}{511}=2.0059$

The ratios tend to approach 2.

c) Since the ratios tend to approach 2, we can deduce that the general term must contain $2^n$ term.

If $t_n=2^n$, then we should have the following sequence

$2,;4,;8,;16,;32,;64,;128,;256,;512;1024$

but the actual sequence is

$1,;5,;7,;17,;31,;65,;127,;257,;511,;1025$

Observe that if $n$ is odd, $t_n=2^n-1$ and if $n$ is even, $t_n=2^n+1$

Therefore, we can use $(-1)^n$ to incorporate this behavior resulting in

$t_n=2^n+(-1)^n$

a) 1, 5, 7, 17, 31, 65, 127, 257, 511, 1025

b) 5, 1.4, 2.43, 1.82, 2.10, 1.95, 2.02, 1.99, 2.00 , it gets closer to 2

c) $t_n=2^n+(-1)^n$