Question

Find the first four terms of the sequence defined by $a_1=3$ and $a_n=3a_{n-1}+2$, for all n > 1.

Answer 1

$a_n=3a_{n-1}+2$ for $n>1$

$\therefore a_2=3a_{2-1}+2=3a_1+2$

$=3\left(3\right)+2=11$ $[\because a_1=3]$

$a_3=3a_{3-1}+2=3a_2+2$

$=\left(11\right)+2=35$ $[\because a_2=11]$

$a_4=3a_{4-1}+2=3a_2+2$

$=3\left(35\right)+2=107$ $[\because a_3=35]$

$\therefore$ The first four terms of A.P. are 3, 11, 35, 107.