Question

Please help meeeeeeeeeeeeeeeeeeeeeeeeeeee

Answer 1

$\\\text{Given that,}~~a_1 =-3\\\\a_n = 3a_{n-1} +5 \\\\a_2 = 3a_{2-1} +5 = 3a_1 +5 = 3(-3) +5 = -9 +5 =-4\\\\a_3 = 3a_{3-1} +5 =3a_2 +5 = 3(-4) +5 = -12 +5 =-7\\\\a_4 = 3a_{4-1} +5 = 3a_3 +5 = 3(-7) +5 = -21+5=-16\\\\\\-3,-4,-7,-16....$

Answer 2

Ok don't hurt yourself I'm here ...

$a(n) = 3a(n - 1) + 5$

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n = 2

put 2 instead of any "n" you can see in the above equation

$a(2) = 3a(2 - 1) + 5$

$a(2) = 3a(1) + 5$

Oh can you see that " a(1) " we know what it is :

$a(1) = - 3$

So :

$a(2) = 3 \times ( - 3) + 5$

$a(2) = - 9 + 5$

$a(2) = - 4$

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n = 3

put 3 instead of blablabla...

$a(3) = 3a(3 - 1) + 5$

$a(3) = 3a(2) + 5$

Oh we just found a(2) so :

$a(3) = 3( - 4) + 5$

$a(3) = - 12 + 5$

$a(3) = - 7$

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n = 4

put 4 ...

$a(4) = 3a(4 - 1) + 5$

$a(4) = 3a(3) + 5$huh a(3)

$a(4) = 3( - 7) + 5$

$a(4) = - 21 + 5$

$a(4) = - 16$

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So the first four term of the sequence is :

$- 3 \: , \: - 4 \: , \: - 7 \: , \: - 16$

There u go...

Have a great day ♡