Question

Nth term of this quadratic sequence -1,2,7,14,23

Answer 1

Since we know it's quadratic, the n-th term will follow the pattern

$x_n = an^2 + bn + c$

for some unknown coefficients a, b, and c.

Given that $x_1=-1$, $x_2=2$, and $x_3=7$, we have the following conditions on these coefficients:

$\begin{cases} a + b + c = -1 \\ 4a + 2b + c = 2 \\ 9a + 3b + c = 7 \end{cases}$

Solve this system to get a = 1, b = 0, and c = -2. Then

$\boxed{x_n = n^2 - 2}$

To solve the system, use elimination.

$(4a + 2b + c) - (a + b + c) = 2 - (-1) \implies 3a + b = 3$

$(9a + 3b + c) - (a + b + c) = 7 - (-1) \implies 8a + 2b = 8 \implies 4a + b = 4$

$(4a + b) - (3a + b) = 4 - 3 \implies a = 1 \implies b = 0 \implies c = -2$