If [tex]v_n[/tex] is the n-th term in the sequence, observe that
[tex]v_2 - v_1 = -26 - (-29) = 3[/tex]
[tex]v_3 - v_2 = -21 - (-26) = 5[/tex]
[tex]v_4 - v_3 = -14 - (-21) = 7[/tex]
[tex]v_5 - v_4 = -5 - (-14) = 9[/tex]
and if the pattern continues,
[tex]v_n - v_{n-1} = 2n - 1[/tex]
so the sequence is defined recursively by
[tex]\begin{cases} v_1 = -29 \\ v_n = v_{n-1} + 2n - 1 & \text{for } n > 1 \end{cases}[/tex]
By this definition,
[tex]v_{n-1} = v_{n-2} + 2(n-1) - 1 = v_{n-2} + 2n - 3[/tex]
[tex]v_{n-2} = v_{n-3} + 2(n-2) - 1 = v_{n-3} + 2n - 5[/tex]
and so on. Then by substitution, we have
[tex]v_n = v_{n-1} + 2n - 1[/tex]
[tex]v_n = (v_{n-2} + 2n - 3) + 2n - 1 = v_{n-2} + 2\times2n - (1 + 3)[/tex]
[tex]v_n = (v_{n-3} + 2n-5) + 2\times2n - (1 + 3) = v_{n-3} + 3\times2n - (1 + 3 + 5)[/tex]
and if we keep doing this we'll eventually get [tex]v_n[/tex] in terms of [tex]v_1[/tex] to be
[tex]v_n = v_1 + (n-1)\times2n - (1 + 3 + 5 + \cdots + (2(n-1)-1))[/tex]
Evaluate the sum:
Let
[tex]S = 1 + 3 + 5 + \cdots + (2(n-1)-1) = 1 + 3 + 5 + \cdots + (2n-3)[/tex]
[tex]S' = 2 + 4 + 6 + \cdots + (2n - 2)[/tex]
Then
[tex]S + S' = 1 + 2 + 3 + 4 + \cdots + (2n-3) + (2n-2) = \displaystyle \sum_{k=1}^{2n-2} k[/tex]
Recall that
[tex]\displaystyle \sum_{i=1}^n i = 1 + 2 + 3 + \cdots + n = \dfrac{n(n+1)}2[/tex]
so that
[tex]S + S' = \dfrac{(2n-2)(2n-1)}2[/tex]
and
[tex]S' = 2 + 4 + 6 + \cdots + (2n-2) = 2 \left(1 + 2 + 3+ \cdots + (n-1)\right) = (n-1)n[/tex]
So, we find
[tex]S = (S + S') - S' = \dfrac{(2n-2)(2n-1)}2 - n(n-1) = (n-1)^2[/tex]
Then the n-th term to the sequence is
[tex]v_n = v_1 +2n(n-1) - S = -29 + 2n^2 - 2n - (n-1)^2 = \boxed{n^2-30}[/tex]
