Question

WILL GIVE BRAINLIESTa(down)1 = 2a(down)n=-3a(down)n-1 + 2

Answer 1

It looks like you're given a recursive sequence $\{a_n\}$ defined by

$\begin{cases}a_1 = 2 \\ a_n = -3a_{n-1} + 2 & \text{for }n>1 \end{cases}$

and I assume you wish to solve it.

By definition of the n-th term $a_n$, we have

$a_2 = -3a_1 + 2$

$a_3 = -3a_2+2 = -3(-3a_1+2)+2 = (-3)^2a_1 - 3\times2 + 2$

$a_4 = -3a_3+2 = -3((-3)^2a_1 - 3\times2 + 2) + 2 = (-3)^3a_1 +(-3)^2\times2-3\times2+2$

and so on.

Notice the patterns:

• the exponent of -3 in the coefficient of $a_1$ on the right side is always 1 less than the subscript on the left side

• the sum of terms not involving $a_1$ is a geometric sum of (subscript - 1) terms,

$2 = 2 \times(-3)^0$

$-3\times2+2 = 2 ((-3)^0 + (-3)^1)$

$(-3)^2\times2-3\times2+2 = 2 ((-3)^0 + (-3)^1 + (-3)^2)$

This suggests that the general n-th term is

$a_n = (-3)^{n-1}a_1 + 2((-3)^0 + (-3)^1 + (-3)^2 + \cdots + (-3)^{n-2})$

and with $a_1=2$, this becomes

$a_n = 2(-3)^{n-1} + 2((-3)^0 + (-3)^1 + (-3)^2 + \cdots + (-3)^{n-2})$

$a_n = 2((-3)^0 + (-3)^1 + (-3)^2 + \cdots + (-3)^{n-2} + (-3)^{n-1})$

or more compactly in sigma notation,

$a_n = 2 \displaystyle \sum_{k=0}^{n-1} (-3)^k$

We can go on to get a closed form for $a_n$. Let S denote the sum,

$S = (-3)^0 + (-3)^1 + (-3)^2 + \cdots + (-3)^{n-2} + (-3)^{n-1}$

Multiply both sides by -3 :

$-3S = (-3)^1 + (-3)^2 + (-3)^3 + \cdots + (-3)^{n-1} + (-3)^{n}$

Subtract this from S - lots of terms will cancel:

$S - (-3S) = (-3)^0 - (-3)^{n}$

$S + 3S = 1 - (-3)^n$

$4S = 1 - (-3)^n$

$S = \dfrac{1 - (-3)^n}4$

It follows that the n-th term of the sequence is explicitly given by

$a_n = 2S = \boxed{\dfrac{1 - (-3)^n}2}$

Edit: Just noticed your comment. You can get the 5th term by plugging the 4th term back into the recurrence:

$a_1 = 2$

$a_2 = -3a_1 + 2 = -3\times2 + 2 = -4$

$a_3 = -3a_2 + 2 = -3\times(-4) + 2 = 14$

$a_4 = -3a_3 + 2 = -3\times14 + 2 = -40$

$a_5 = -3a_4 + 2 = -3\times(-40) + 2 = \boxed{122}$

Or, using the n-th term we found above,

$a_5 = \dfrac{1 - (-3)^5}2 = \dfrac{1 - (-243)}2 = \dfrac{244}2 = \boxed{122}$