Answer
$(a)$ 2, 12, 72, 432, 2592
$(b)$ 2, 4, 16, 256, 65536
$(c)$ 1, 2, 5, 11, 26
$(d)$ 1, 1, 6, 27, 204
$(e)$ 1, 2, 0, 1, 3
Work Step by Step
$(a)$
Here, $a_{n}$ = $6a_{n-1}$; $a_0$ = 2
$a_1$ = $6a_0$ = $6\times2$ = $12$
$a_2$ = $6a_1$ = $6\times12$ = $72$
$a_3$ = $6a_2$ = $6\times72$ = $432$
$a_4$ = $6a_3$ = $6\times432$ = $2592$
So the first five terms of the sequence is 2, 12, 72, 432, 2592
$(b)$
Here, $a_n$ = $a_{n-1}^{2}$ ; $a_1$ = 2
$a_2$ = $a_{1}^2$ = $2^2$ = $4$
$a_3$ = $a_{2}^2$ = $4^2$ = $16$
$a_4$ = $a_{3}^2$ = $16^2$ = $256$
$a_5$ = $a_{4}^2$ = $256^2$ = $65536$
so the first five terms of the sequence is 2, 4, 16, 256, 65536.
$(c)$
Here, $a_n = a_{n-1} + 3a_{n-2} ; a_0 = 1; a_1 = 2 $
$a_2 = a_1 + 3a_0 = 2 + (3\times1) = 2 + 3 = 5 $
$a_3 = a_2 + 3a_1 = 5 + (3\times2) = 5 + 6 = 11 $
$a_4 = a_3 + 3a_2 = 11 + (3\times5) = 11 + 15 = 26 $
so the first five terms of the sequence is 1, 2, 5, 11, 26.
$(d)$
$a_n = na_{n-1} + n^2a_{n-2} , a_0 = 1, a_1 = 1$
$a_2 = 2a_1 + 2^2a_0 = (2\times1) + (4\times1) = 2 + 4 = 6$
$a_3 = 3a_2 + 3^2a_1 = (3\times6) + (9\times1) = 18 + 9 = 27$
$a_4 = 4a_3 + 4^2a_2 = (4\times27) + (16\times6) = 108 + 96 = 204$
so the first five terms of the sequence is 1, 1, 6, 27, 108
$(e)$
$a_n = a_{n-1} + a_{n-3} ; a_0 = 1, a_1 = 2, a_2 = 0$
$a_3 = a_2 + a_0 = 0 + 1 = 1$
$a_4 = a_3 + a_1 = 1 + 2 = 3$
so the first five terms of the sequence is 1, 2, 0, 1, 3
