The first four terms of a sequence are given by $v_1 = -9$, $v_2 = 7$, $v_3 = 47$ and $v_4 = 141$. It is given that $v_n$ is a cubic polynomial in $n$. Find $v_n$ in terms of $n$.

1. Formulate a system of linear equations by substituting the given terms into the general form of a cubic polynomial.

Let the cubic polynomial be $v_n = an^3 + bn^2 + cn + d$.
Substitute $n = 1, 2, 3, 4$ and the corresponding values of $v_n$:
$a(1)^3 + b(1)^2 + c(1) + d = -9 \implies a + b + c + d = -9 \quad \text{--- (1)}$
$a(2)^3 + b(2)^2 + c(2) + d = 7 \implies 8a + 4b + 2c + d = 7 \quad \text{--- (2)}$
$a(3)^3 + b(3)^2 + c(3) + d = 47 \implies 27a + 9b + 3c + d = 47 \quad \text{--- (3)}$
$a(4)^3 + b(4)^2 + c(4) + d = 141 \implies 64a + 16b + 4c + d = 141 \quad \text{--- (4)}$

2. Solve the system of equations simultaneously to find the coefficients.

Subtract (1) from (2), (3), and (4) to eliminate $d$:
$7a + 3b + c = 16 \quad \text{--- (5)}$
$26a + 8b + 2c = 56 \implies 13a + 4b + c = 28 \quad \text{--- (6)}$
$63a + 15b + 3c = 150 \implies 21a + 5b + c = 50 \quad \text{--- (7)}$

Subtract (5) from (6) and (7) to eliminate $c$:
$6a + b = 12 \quad \text{--- (8)}$
$14a + 2b = 34 \implies 7a + b = 17 \quad \text{--- (9)}$

Subtract (8) from (9):
$a = 5$

Substitute $a = 5$ into (8):
$6(5) + b = 12 \implies b = -18$

Substitute $a = 5, b = -18$ into (5):
$7(5) + 3(-18) + c = 16 \implies c = 35$

Substitute $a = 5, b = -18, c = 35$ into (1):
$5 - 18 + 35 + d = -9 \implies d = -31$

3. State the final expression for $v_n$ in terms of $n$.

Substitute the found coefficients back into the general polynomial:
$v_n = 5n^3 - 18n^2 + 35n - 31$