Answer
$(a)\space -1, 2, -4, 8, -16, 32 $
$(b) \space 2, -1, -3, -2, 1, 3$
$(c) \space 1, 3, 27, 2187, 14348907, (3\times14348907^2)$
$(d) \space -1, 0, 1, 3, 13, 74$
$(e)\space 1, 1, 2, 2, 1, 1$
Work Step by Step
$(a)$
Here, $a_n = -2a_{n-1} ;\space\space a_0 = -1$
$a_1 = -2a_0 \space= -2\times-1 \space = 2$
$a_2 = -2a_1 \space= -2\times2 \space = -4$
$a_3 = -2a_2 \space= -2\times-4 \space = 8$
$a_4 = -2a_3 \space= -2\times8 \space = -16$
$a_5 = -2a_4 \space= -2\times-16 \space = 32$
so the first six terms of the sequence is -1, 2, -4, 8, -16, 32.
$(b)$
Here, $a_n = a_{n-1} - a_{n-2}; \space a_0 = 2; \space a_1 = -1$
$a_2 = a_1 - a_0 \space = -1 - 2 \space = -3 $
$a_3 = a_2 - a_1 \space = -3 - (-1) \space = -2 $
$a_4 = a_3 - a_2 \space = -2 - (-3) \space = 1 $
$a_5 = a_4 - a_3 \space = 1 - (-2) \space = 3 $
so the first six terms of the sequence is 2, -1, -3, -2, 1, 3.
$(c)$
Here, $a_n = 3a_{n-1}^2 \space =; a_0 = 1$
$a_1 = 3a_0^2 \space = 3\times(1^2) \space = 3\times(1) = 3$
$a_2 = 3a_1^2 \space = 3\times(3^2) \space = 3\times(9) = 27$
$a_3 = 3a_2^2 \space = 3\times(27^2) \space = 3\times(729) = 2187$
$a_4 = 3a_3^2 \space = 3\times(2187^2) \space = 3\times(4782969) = 14348907$
$a_5 = 3a_4^2 \space = 3\times(14348907^2)$
so the first six terms of the sequence is 1, 3, 27, 2187, 14348907, ($3\times14348907^2$).
$(d)$
Here, $a_n = na_{n-1} + a_{n-2}^2; \space a_0 = -1; \space a_1 = 0$
$a_2 = 2a_1 + a_0^2 \space = 2(0) + (-1)^2 \space = 0 + 1 \space= 1$
$a_3 = 3a_2 + a_1^2 \space = 3(1) + (0)^2 \space = 3 + 0 \space= 3$
$a_4 = 4a_3 + a_2^2 \space = 4(3) + (1)^2 \space = 12 + 1 \space= 13$
$a_5 = 5a_4 + a_3^2 \space = 5(13) + (3)^2 \space = 65 + 9 \space= 74$
so the first six terms of the sequence is -1, 0, 1, 3, 13, 74.
$(e)$
Here, $a_n = a_{n-1} - a_{n-2} + a_{n-3} \space ; a_0 = 1; \space ; a_1 = 1; \space a_2 = 2$
$a_3 = a_2 - a_1 + a_0 \space = 2 -1+1 = 2 $
$a_4 = a_3 - a_2 + a_1 \space = 2 -2+1 = 1 $
$a_5 = a_4 - a_3 + a_2 \space = 1 -2+2 = 1 $
so the first six terms of the sequence is 1, 1, 2, 2, 1, 1.
