It looks like you're given a recursive sequence [tex]\{a_n\}[/tex] defined by
[tex]\begin{cases}a_1 = 2 \\ a_n = -3a_{n-1} + 2 & \text{for }n>1 \end{cases}[/tex]
and I assume you wish to solve it.
By definition of the n-th term [tex]a_n[/tex], we have
[tex]a_2 = -3a_1 + 2[/tex]
[tex]a_3 = -3a_2+2 = -3(-3a_1+2)+2 = (-3)^2a_1 - 3\times2 + 2[/tex]
[tex]a_4 = -3a_3+2 = -3((-3)^2a_1 - 3\times2 + 2) + 2 = (-3)^3a_1 +(-3)^2\times2-3\times2+2[/tex]
and so on.
Notice the patterns:
• the exponent of -3 in the coefficient of [tex]a_1[/tex] on the right side is always 1 less than the subscript on the left side
• the sum of terms not involving [tex]a_1[/tex] is a geometric sum of (subscript - 1) terms,
[tex]2 = 2 \times(-3)^0[/tex]
[tex]-3\times2+2 = 2 ((-3)^0 + (-3)^1)[/tex]
[tex](-3)^2\times2-3\times2+2 = 2 ((-3)^0 + (-3)^1 + (-3)^2)[/tex]
This suggests that the general n-th term is
[tex]a_n = (-3)^{n-1}a_1 + 2((-3)^0 + (-3)^1 + (-3)^2 + \cdots + (-3)^{n-2})[/tex]
and with [tex]a_1=2[/tex], this becomes
[tex]a_n = 2(-3)^{n-1} + 2((-3)^0 + (-3)^1 + (-3)^2 + \cdots + (-3)^{n-2})[/tex]
[tex]a_n = 2((-3)^0 + (-3)^1 + (-3)^2 + \cdots + (-3)^{n-2} + (-3)^{n-1})[/tex]
or more compactly in sigma notation,
[tex]a_n = 2 \displaystyle \sum_{k=0}^{n-1} (-3)^k[/tex]
We can go on to get a closed form for [tex]a_n[/tex]. Let S denote the sum,
[tex]S = (-3)^0 + (-3)^1 + (-3)^2 + \cdots + (-3)^{n-2} + (-3)^{n-1}[/tex]
Multiply both sides by -3 :
[tex]-3S = (-3)^1 + (-3)^2 + (-3)^3 + \cdots + (-3)^{n-1} + (-3)^{n}[/tex]
Subtract this from S - lots of terms will cancel:
[tex]S - (-3S) = (-3)^0 - (-3)^{n}[/tex]
[tex]S + 3S = 1 - (-3)^n[/tex]
[tex]4S = 1 - (-3)^n[/tex]
[tex]S = \dfrac{1 - (-3)^n}4[/tex]
It follows that the n-th term of the sequence is explicitly given by
[tex]a_n = 2S = \boxed{\dfrac{1 - (-3)^n}2}[/tex]
Edit: Just noticed your comment. You can get the 5th term by plugging the 4th term back into the recurrence:
[tex]a_1 = 2[/tex]
[tex]a_2 = -3a_1 + 2 = -3\times2 + 2 = -4[/tex]
[tex]a_3 = -3a_2 + 2 = -3\times(-4) + 2 = 14[/tex]
[tex]a_4 = -3a_3 + 2 = -3\times14 + 2 = -40[/tex]
[tex]a_5 = -3a_4 + 2 = -3\times(-40) + 2 = \boxed{122}[/tex]
Or, using the n-th term we found above,
[tex]a_5 = \dfrac{1 - (-3)^5}2 = \dfrac{1 - (-243)}2 = \dfrac{244}2 = \boxed{122}[/tex]
