Notice that the difference in the absolute values of consecutive coefficients is constant:
|-7| - 1 = 6
13 - |-7| = 6
|-19| - 13 = 6
and so on. This means the coefficients in the given series
[tex]\displaystyle \sum_{i=1}^\infty c_i a^{i-1} = \sum_{i=1}^\infty |c_i| (-a)^{i-1} = 1 - 7a + 13a^2 - 19a^3 + \cdots[/tex]
occur in arithmetic progression; in particular, we have first value [tex]c_1 = 1[/tex] and for [tex]n>1[/tex], [tex]|c_i|=|c_{i-1}|+6[/tex]. Solving this recurrence, we end up with
[tex]|c_i| = |c_1| + 6(i-1) \implies |c_i| = 6i - 5[/tex]
So, the sum to [tex]n[/tex] terms of this series is
[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \underbrace{\sum_{i=1}^n i (-a)^{i-1}}_{S'} - 5 \underbrace{\sum_{i=1}^n (-a)^{i-1}}_S[/tex]
The second sum [tex]S[/tex] is a standard geometric series, which is easy to compute:
[tex]S = 1 - a + a^2 - a^3 + \cdots + (-a)^{n-1}[/tex]
Multiply both sides by [tex]-a[/tex] :
[tex]-aS = -a + a^2 - a^3 + a^4 - \cdots + (-a)^n[/tex]
Subtract this from [tex]S[/tex] to eliminate the intermediate terms to end up with
[tex]S - (-aS) = 1 - (-a)^n \implies (1-(-a)) S = 1 - (-a)^n \implies S = \dfrac{1 - (-a)^n}{1 + a}[/tex]
The first sum [tex]S'[/tex] can be handled with simple algebraic manipulation.
[tex]S' = \displaystyle \sum_{i=1}^n i (-a)^{i-1}[/tex]
[tex]\displaystyle S' = \sum_{i=0}^{n-1} (i+1) (-a)^i[/tex]
[tex]\displaystyle S' = \sum_{i=0}^{n-1} i (-a)^i + \sum_{i=0}^{n-1} (-a)^i[/tex]
[tex]\displaystyle S' = \sum_{i=1}^{n-1} i (-a)^i + \sum_{i=1}^n (-a)^{i-1}[/tex]
[tex]\displaystyle S' = \sum_{i=1}^n i (-a)^i - n (-a)^n + S[/tex]
[tex]\displaystyle S' = -a \sum_{i=1}^n i (-a)^{i-1} - n (-a)^n + S[/tex]
[tex]\displaystyle S' = -a S' - n (-a)^n + \dfrac{1 - (-a)^n}{1 + a}[/tex]
[tex]\displaystyle (1 + a) S' = \dfrac{1 - (-a)^n - n (1 + a) (-a)^n}{1 + a}[/tex]
[tex]\displaystyle S' = \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2}[/tex]
Putting everything together, we have
[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 S' - 5 S[/tex]
[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} = 6 \dfrac{1 - (n+1)(-a)^n + n (-a)^{n+1}}{(1+a)^2} - 5 \dfrac{1 - (-a)^n}{1 + a}[/tex]
[tex]\displaystyle \sum_{i=1}^n (6i-5) (-a)^{i-1} =\boxed{\dfrac{1 - 5a - (6n+1) (-a)^n + (6n-5) (-a)^{n+1}}{(1+a)^2}}[/tex]
