Is there a general formula for the sum of a quadratic sequence?

Question

I tried Googling "formula for sum of quadratic sequence", which did not give me anything useful. I just want an explicit formula for figuring out a sum for a quadratic sequence. For example, how would you figure out the sum of $2+6+12+20+\dots+210$? Can someone please help? Thanks

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For those of you who do not know, a quadratic sequence is a sequence where the differences of the differences between the terms are constant. Let's use $2+6+12+20+\dots$ as an example. The differences between the terms are $4$, $6$, $8$, etc. The difference between the differences of the terms is $2$. So the sequence will continue like $2+6+12+20+30+42+56+72+\dots$

Answer 1

Yes there is. Ever wonder why this is called quadratic sequence? Quadratic refers to squares right? This is just constant difference of difference. So where's the connection? Well as it turns out, all terms of a quadratic sequence are expressible by a quadratic polynomial. What do I mean? Consider this

$$ t_n = n+n^2 $$

Subsituiting $n=1,2,3,\cdots$ generates your terms. By the way, $202$ doesn't occur in this sequence, the 13th term is $182$ and the $14th$ term is $210$. I am assuming it was supposed to be $210$.

So we need to find

$$ \sum_{i=1}^{n}i+i^2 = \sum_{i=1}^{n}i+\sum_{i=1}^{n}i^2 $$

where $n=14$. There are well known formulas for $\sum_{i=1}^{n}i$ and for $\sum_{i=1}^{n}i^2$. Substituting them, we get,

$$\frac{n(n+1)}{2} + \frac{n(n+1)(2n+1)}{6}$$ $$=\frac{n(n+1)}{2}\left(1+\frac{2n+1}{3}\right)$$ $$=\frac{n(n+1)}{2}\left(\frac{3+2n+1}{3}\right)$$ $$=\frac{n(n+1)}{2}\left(\frac{2n+4}{3}\right)$$ $$=\frac{n(n+1)(n+2)}{3}$$

where $n=14$. Thus our sum is $1120$.

Answer 2

For the general case, N terms are to be summed

$$S=a_0+a_1+a_2+...+a_{N-1}$$

The formula for the n-th term is $$a_n=a_0+(a_1-a_0)n+(a_2-2a_1+a_0)\frac{n(n-1)}{2}$$

Using the results

$$ \sum_{n=0}^{N-1}n=\frac{N(N-1)}{2}$$

and

$$ \sum_{n=0}^{N-1}n(n-1)=\frac{N(N-1)(N-2)}{3}$$

leads to

$$S=\sum_{n=0}^{N-1}a_n=a_0N+(a_1-a_0)\frac{N(N-1)}{2}+(a_2-2a_1+a_0)\frac{N(N-1)(N-2)}{6}$$

Answer 3

Although this may not be needed as of now; but I thought about sums of quadratic sequences myself and I managed to derive a general formula for it, so I might as well post it here:

Where is the number of terms to compute, is the starting term, is the first difference and is the constant difference between the differences. I use the subscript to denote that it is the quadratic sequence sum function.

Answer 4

In this particular case, $$\begin{align} &2+6+12+20+\cdots+210\\\\ &=2(1+3+6+10+\cdots+105)\\\\ &=2\left[\binom 22 +\binom 32 +\binom 42+\binom 52+\cdots \binom {15}2\right]\\ &=2\sum_{r=2}^{15}\binom r2\\ &=2\binom {16}3\\ &=\color{red}{1120}\qquad\blacksquare \end{align}$$

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This is the sum of triangular numbers (where the difference of the difference is constant) and the result is a pyramidal number (all scaled by 2). The summation can be shown as

$$\begin{align} &2\cdot (1)+\\ &2\cdot (1+2)+\\ &2\cdot (1+2+3)+\\ &2\cdot (1+2+3+4)+\\ &\quad \vdots\qquad\qquad\qquad\ddots\\ &2\cdot (1+2+3+\cdots+15) \end{align}$$

If we tried to sum a series where the difference of the difference of the difference is constant, i.e. sum of pyramidal numbers, the result would be a pentatope number. And so on...

An example of the summation of pyramidal numbers, extending from the original question, would be

$$2+8+20+40+\cdots+910 =2 \sum_{r=1}^{15} \binom r3=2 \binom {16}4=1820\\$$

Answer 5

I figured out the below way of doing it just know at one o'clock right before bedtime, so if it is faulty than that is my mistake.

Any series has a certain term-to-term rule. For a quadratic that term-to-term rule is in the form

$ax^2+bx+c$

The series will simply be that term-to-term rule with $x$ replaced by $0$, then by $1$ and so on. This can be written as

$\sum_{x=0}^p ax^2+bx+c$

We already know that

$\sum_{x=0}^p x=\frac{x(x+1)}{2}$

$\sum_{x=0}^p x^2=\frac{x(x+1)(2x+1)}{6}$

$\sum_{x=0}^p c=c*p$

These can all be proved easily and you can find said proofs online. Because addition is commutative (the order does not matter) the quadratic sum above rewritten as

$a*\sum_{x=0}^p x^2+b*\sum_{x=0}^p x + \sum_{x=0}^p c$

using the above identities, this can be simplified to

$a*\frac{p(p+1)(2p+1)}{6}+b*\frac{p(p+1)}{2}+c*p$

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In your example the term-to-term rule is

$1x^2+1x+0$

And we want to find the sum until the term that will give us $210$. This number can be calculated to be $14$. That means that $14^2+14=210$. So in our formula

$p=14$

$a=1$,

$b=1$

$c=0$.

Plug that in and your sum is $980$

Answer 6

I have a sum formula for quadratic equation.

$$s=\frac{n}{6} (3d(n-1)+(n-1)(n-2)c) +an $$

Where $n$ is the number of terms, $d$ is the first difference, $c$ is the constant difference or difference of difference, $a$ is first term. For calculation of sum first put value of $a$ $d$ $c$ then you get a formula like $an^2+bn+c$.