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Find the nth term of this quadratic sequence. 3,11,25,45

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Since you know the sequence is quadratic, we can find constants a, b, and c such that the n-th term of the sequence (denoted by f(n)) is


f(n) = an^2 + bn + c

We're given f(1) = 3, f(2) = 11, and f(3) = 25. (We have three unknowns, so we only need three conditions.) Solve for the constants:


\begin{cases}f(1) = a + b + c = 3 \\ f(2) = 4a + 2b + c = 11 \\ f(3) = 9a + 3n + c = 25 \end{cases} \implies a=3, b=-1, c=1

Then the n-th term of the sequence is


f(n) = \boxed{3n^2 - n + 1}

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