Question

The first five terms of a quadratic sequence are 3, 12, 25, 42, 63.
Find the nth term of this sequence.

Answer 1

Since we know the sequence is quadratic, we expect the $n$-th term to have the general form

$x_n = an^2 + bn + c$

Plug in the first 3 known values of the sequence to form a system of equations.

$x_1 = 3 = a + b + c$

$x_2 = 12 = 4a + 2b + c$

$x_3 = 25 = 9a + 3b + c$

Eliminating $c$ gives

$(4a + 2b + c) - (a + b + c) = 12 - 3 \implies 3a + b = 9$

$(9a + 3b + c) - (a + b + c) = 25 - 3 \implies 8a + 2b = 22 \implies 4a + b = 11$

Eliminating $b$ gives

$(4a + b) - (3a + b) = 11 - 9 \implies a = 2$

Solving for $b,c$, we get

$4a + b = 11 \implies b = 11-4\cdot2 = 3$

$a+b+c=3 \implies c=3-2-3 = -2$

So, the $n$-th term of the sequence is

$x_n = \boxed{2n^2 + 3n - 2}$