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Given the quadratic sequence $$f(n)=1, 7, 19, 37, \cdots$$

To calculate the $f(n)$ for $n\ge1$. $$f(n)=an^2+bn+c$$

We start with the general quadratic function, then sub in for $n:=1,2$ and $3$

$$f(1)=a+b+c$$ $$f(2)=4a+2b+c$$ $$f(3)=9a+3b+c$$

Now solve the simultaneous equations

$$a+b+c=1\tag1$$ $$4a+2b+c=7\tag2$$ $$9a+3b+c=19\tag3$$

$(2)-(1)$ and $(3)-(2)$

$$3a+b=6\tag4$$ $$5a+b=12\tag5$$

$(5)-(4)$ $$a=3$$ $$b=-3$$ $$c=1$$

$$f(n)=3n^2-3n+1$$

This method is very long. Is there another easy of calculating the $f(n)$?

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    Yes: just do what you did, but don't substitute the numbers; instead, keep $f(1), f(2), f(3)$ throughout. Then you end up with a fully general formula. – Patrick Stevens Mar 28 '18 at 11:13

4 Answers4

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Yes!

You know $f(1)= 1 \implies f(1) - 1 = 0 \implies f(x) - 1$ has a root at x=1

Now,

$f(x) - 1 = a(x-1)(x-b)$

Put $x = 2$, $a(2-b) = 6$

Put x = 3, $a*2*(3-b) = 18$

Divide both equation, we get $b = 0 $ and a = 3

$f(x) - 1 = 3*(x-1)x \implies f(x) = 3x^2 - 3x + 1$

kayush
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4

Another standard way is to calculate a difference scheme and then to work backwards: $$\begin{matrix} 0 & & 1 & & 2 & & 3 & & 4 \\ & & 1 && 7 && 19 && 37 \\ && &6& & 12 && 18 & \\ &&&& 6 && 6 && \\ \end{matrix} \Rightarrow \begin{matrix} & 0 & & 1 & & 2 & & 3 & & 4 \\ \color{blue}{c}= &\color{blue}{1} & & 1 && 7 && 19 && 37 \\ \color{blue}{a+b}= & &\color{blue}{0}&&6& & 12 && 18 & \\ \color{blue}{2a}= &&& \color{blue}{6}&& 6 && 6 && \\ \end{matrix}$$ $$\Rightarrow a = 3, \; b= -3, \; c = 1 \Rightarrow f(n) = 3n^2-3n+1$$

trancelocation
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Take a better basis. Namely, $\{(n-1)(n-2),(n-1)(n-3),(n-2)(n-3)\}$. If $$f(n) = \alpha(n-1)(n-2) + \beta(n-1)(n-3) + \gamma(n-2)(n-3),$$ then: $$f(1) = 0 + 0 + \gamma(1-2)(1-3),$$ $$f(2) = 0 + \beta(2-1)(2-3) + 0,$$ $$f(3) = \alpha(3-1)(3-2) + 0 + 0.$$

Martín-Blas Pérez Pinilla
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A better explained version of @trancelocation's answer:

$$\begin{array}{} 1 & & 7 & & 19 & & 37 \\ & 6 & & 12& &18 & \\ & & 6 & & 6 & & \\ \end{array}$$

The second row has equation $6n$. Therefore, $\big(a(n+1)^2+b(n+1)+c \big) - \big(an^2+bn+c \big)$ $ = 6n$, and so:

$$\big(a(n^2+2n+1)+b(n+1)+c \big) - \big(an^2+bn+c \big) = 6n$$ $$(2n+1)a + b = 6n$$

$a=3$ gives $6n+3$, and so $b=-3$. 'Plugging' $n=1$ into $3n^2-3n$ gives $0$, $c=1$, which gives $3n^2-3n+1$.

Toby Mak
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