Question

. Find the first five terms of the sequence given by a1 = 2, an = 3an-1 - 1

Answer 1

Answer:

Given the sequence:

$a_1 =2$

$a_n = 3a_{n-1}-1$

where, n is the number of terms.

We have to find the first five terms of the sequence.

For n = 2;

$a_2 = 3a_{2-1}-1$

⇒$a_2 = 3a_{1}-1$

Substitute the given values we have;

$a_2 = 3 \cdot 2 -1 = 6-1 = 5$

For n = 3

$a_3 = 3a_{3-1}-1$

⇒$a_3= 3a_{2}-1$

Substitute the given values we have;

$a_3 = 3 \cdot 5 -1 = 15-1 = 14$

For n = 4

$a_4 = 3a_{4-1}-1$

⇒$a_4= 3a_{3}-1$

Substitute the given values we have;

$a_4 = 3 \cdot 14 -1 = 42-1 = 41$

For n = 5

$a_5 = 3a_{5-1}-1$

⇒$a_5= 3a_{4}-1$

Substitute the given values we have;

$a_5= 3 \cdot 41 -1 = 123-1 = 122$

Therefore, the first five terms of the sequence are:

$a_1=2$, $a_2=5$, $a_3 =14$,  $a_4 =41$ and  $a_5=122$,

Answer 2

An = 3 *(an-1) - 1

a1 = 2
a2 = 3* (2) - 1 = 6 - 1 = 5
a3 = 3* (5) - 1 = 15 -1 = 14
a4 = 3* (14) - 1 = 42 -1 = 41
a5 = 3* (41) - 1 = 123 -1 = 122