Why are Quadratic progressions called so and how is general term of a quadratic sequence $an^2 + bn + c$?

Question

When I went through various sources on the Internet, about Quadratic Progression, all of them said that a general term in a quadratic sequence would be of the form $an^2 + bn + c$, and explained how to find the $n_{th}$ term of specific sequences like $1,4,9,16,...$ and $5,7,11,17,...$ etc. But none of them really explained how that expression is obtained or why it works.

Some general results I found where;

Any term of a quadratic sequence will be of form: $$an^2+bn+c$$ and; $$2a=2^{nd} \mathrm{difference (constant)}$$ $$3a+b=2^{nd}\mathrm {term}-1^{st}\mathrm {term}$$ $$a+b+c=1^{st}\mathrm{term}$$

Why are Quadratic Sequences called so, and how do we obtain these relations(conclusions) and why does this work every time?

Answer 1

For the formula, I think you may have the idea backwards. "Quadratic" basically means $an^2+bn+c $ (historically related to things like "a square has four sides" and "quad is the Latin root for 'four'"), so that formula could be treated as true by the definition of "quadratic sequence".

The first-term formula comes from substituting in $n=1$, since $n$ is the variable being used to denote which term we're looking at. $a*1^2+b*1+c=a+b+c $.

The difference between the first two terms comes from writing down the first and second terms and taking their difference: $(a*2^2+b*2+c)-(a*1^2+b*1+c) =a*3+b $

The second difference being referred to is the difference between adjacent differences. For example, with $5,7,11,17,\ldots $, the differences between adjacent terms are $(7-5),(11-7),(17-11),\ldots=2,4,6,\ldots $. Then the second-level differences are $(4-2),(6-4),\ldots$ and happen to always be $2$.

In general, if we start looking for this second difference at the $n^{\text{th}} $ term and the two terms that follow, the two consecutive differences are $\left(a(n+2)^2+b(n+2)+c\right)-\left(a(n+1)^2+b(n+1)+c\right)=a(2n+3)+b $ and $\left(a(n+1)^2+b(n+1)+c\right)-\left(an^2+bn+c\right)=a(2n+1)+b $. Then the second-level difference is $\left(a(2n+3)+b\right)-\left(a(2n+1)+b\right)=2a$.

The second difference is constant because this formula "$2a $" has no $n $ in it, so won't change as we look at different points in the sequence.

Answer 2

Many people would in fact define a quadratic progression by the formula $an^{2} + bn + c$, but if you're interested in how the sequence determines the formula, I'll try to provide an answer (bottom).

Firstly, the name 'quadratic' is typically used in mathematics whenever there is some squared variable (whether it be a quadratic progression, quadratic equation, quadratic number field, quadratic residue, and so on) so the $n^{2}$ term would mean that the name is appropriate. After all, a sequence is just a special type of function --- instead of mapping real numbers to real numbers, a sequence maps the natural numbers to real numbers.

If we were to ask why the sequence $1, 4, 9, 16, 25$ should be called quadratic sequence without the knowledge of the $an^{2} + bn + c$ formula, this follows from the fact that the terms are all squares so there is some quadratic stuff going on.

For a more general quadratic sequence like $5, 7, 11, 17$ where there are no obvious squares, this has the name quadratic since the second differences in the numbers are constant. That is, the difference between the numbers are $2, 4, 6$ (which is linear), and the difference of those numbers is $2, 2$ which is constant. Effectively we have found the rate of change of the numbers (also called finding the derivative), and the second rate of change (or the second derivative) is constant. What type of function gives you something constant when you differentiate it twice? A quadratic one! Since the second differences are constant, this is a quadratic sequence. (Go check that this is also true for the first sequence, so there is no difference really in which viewpoint we use.)

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Now to derive the formula, the sequences always follow the same progression so we'll do a specific case and then you can try to generalise. Consider $$ 5, 7, 11, 17, \ldots $$ The terms can be rewritten $$ 5,\hspace{15pt} 5 + 2,\hspace{15pt} 5 + 2 + 4,\hspace{15pt} 5 + 2 + 4 + 6, \ldots $$ and again as $$ 5,\hspace{10pt} 5 + 1\cdot 2,\hspace{10pt} 5 + 1\cdot 2 + 2\cdot 2,\hspace{10pt} 5 + 1\cdot 2 + 2\cdot 2 + 3\cdot 2, \ldots $$ and finally as $$ 5 + 2\cdot(0),\hspace{10pt} 5 + 2\cdot(0 + 1),\hspace{10pt} 5 + 2\cdot(0 + 1 + 2),\hspace{10pt} 5 + 2\cdot(0 + 1 + 2 + 3), \ldots $$ We can see a pattern. In the first term we have $5 + 2\cdot(0)$, in the second term we have $5 + 2\cdot(0 + 1)$, in the third term we have $5 + 2\cdot(0 + 1 + 2)$, $\ldots$, in the $n^{\text{th}}$ term we have $5 + 2\cdot(0 + 1 + 2 + \ldots + n - 1)$. Now the formula for the sum of the first $n - 1$ positive integers is $$ 0 + 1 + 2 + 3 + \ldots + n - 1 = 0 + \sum_{i=1}^{n-1} = \frac{1}{2}n(n - 1). $$ Thus the $n^{\text{th}}$ term is $$ 5 + 2\cdot(0 + 1 + 2 + \ldots + n - 1) = 5 + 2\cdot\left(\frac{1}{2}n(n-1)\right) = 5 + n(n - 1) = n^{2} - n + 5. $$ This is precisely the form we expected! By subbing the $n$ values in to the above, you can see that this corresponds precisely to the sequence above.

Updated: The general term was found by observing the pattern in the table below. The $n$ corresponds to where in the sequence the term appears (ie, first or second or third and so on).

$$ \begin{array}{r|l} n & \text{Term} \\\hline 1 & 5 + 2\cdot(0)\\ 2 & 5 + 2\cdot(0 + 1) \\ 3 & 5 + 2\cdot(0 + 1 + 2) \\ 4 & 5 + 2\cdot(0 + 1 + 2 + 3) \\ 5 & 5 + 2\cdot(0 + 1 + 2 + 3 + 4) \\ 6 & 5 + 2\cdot(0 + 1 + 2 + 3 + 4 + 5) \\ \end{array} $$

Note that I have written $5$ as $5 + 2\cdot(0)$ which are clearly equal just to make the pattern more clear. We see that in the $n^{\text{th}}$ term we have a sum of all the non-negative integers up to $n-1$, so that the $n^{\text{th}}$ term is $$ 5 + 2\cdot\big(0 + 1 + 2 + 3 + 4 + \ldots + (n - 2) + (n - 1)\big). $$

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It might also be worth pointing out that if you have the formula $c + bn + an^{2}$, then the first few terms are summarised below.

$$ \begin{array}{r|l} n & \text{Term} \\\hline 1 & c + 1\cdot(b + a) \\ 2 & c + 2\cdot(b + 2a) \\ 3 & c + 3\cdot(b + 3a) \\ 4 & c + 4\cdot(b + 4a) \\ 5 & c + 5\cdot(b + 5a) \\ 6 & c + 6\cdot(b + 6a) \\ \end{array} $$