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

User Nicholas Muir
by
2.9k points

1 Answer

23 votes
23 votes

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

User Cory House
by
3.1k points
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