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I have been stuck in finding the general term and the sum of $n$ terms of the series

$$12,\,40,\,90,\,168,\,280,\,432, ...$$

I am not able to see any relationship between the successive terms of the series.

Is there a pattern which I am not able to see?

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    $\begingroup$ There has to be. If there is no pattern, you can't come up with the other terms. $\endgroup$ – Laray Jan 17 '17 at 16:07
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    $\begingroup$ There is a pattern, which clearly points at the next term being $$42$$ as every other term after it. Thus, for every $n\geqslant6$, the sum of the $n$ first terms is $$42n+770$$ $\endgroup$ – Did Jan 17 '17 at 16:12
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    $\begingroup$ If you compute repeated differences, then you'll see that there is a simple cubic polynomials that explains these values: WA. $\endgroup$ – lhf Jan 17 '17 at 16:13
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Clearly there is a pattern.

Firstly find the orders of differences.

They are

28, 50, 78, 112, 152,...

22,     28,     34,     40, ....

    6,      6,      6, ....

        0,      0,.....

Hence, the $n^{th}$ term can be written as

$12+28(n-1)+\frac{22n(n-1)(n-2)}{2!}+\frac{6n(n-1)(n-2)(n-3)}{3!}$

=$n^3+5n^2+6n$

The sum of n terms of the above series is

$\sum {n^3}+5 \sum {n^2}+6\sum {n}$

=$12n$+$\frac{28n(n-1)}{2!}+\frac{22n(n-1)(n-2)}{3!}+\frac{6n(n-1)(n-2)(n-3)}{4!}$

=$\frac{n(3n^2+26n+69n+46}{12})$

=$\frac{1}{12}n(n+1)(3n^2+23n+46)$

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There is indeed. Observe that if you go on taking successive differences (at each step) then, you get

$$12,40,90,168,280,432,...$$ $$28,50,78,112,152,...$$ $$22,28,34,40,...$$ $$6,6,6,6,...$$

Perhaps, a constant sequence. Does this strike something?

Yes! You can assert that, the general term of the given sequence is of the form $x_n = an^3+bn^2+cn+d$. Solve for $(a,b,c,d)$ using the fact that $x_1=12, x_2 = 40, x_3 = 90$ and $x_4 = 168$.

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The Online Encyclopedia of Integer Sequences thinks that the terms are given by $$a_n=n^3+5n^2+6n$$

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    $\begingroup$ So does WA. $\endgroup$ – lhf Jan 17 '17 at 16:14
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    $\begingroup$ Divide the entries by $2$ and you get oeis.org/A005564 $\endgroup$ – Barry Cipra Jan 17 '17 at 16:24
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Per my solution here it can be shown that the $n$-th term ($n=0,1,2,3,\cdots$), $b_n$ is given by

$$b_n=\sum_{r=0}^{\min(3,n)}\binom nr a_r$$ where $a_r=12,28,22,6$ for $r=0,1,2,3$, i.e.

$$\begin{align} b_0&=\binom 00 12&&=12\\ b_1&=\binom 10 12+\binom 11 28&&=40\\ b_2&=\binom 20 12+\binom 21 28+\binom 22 22 &&=90\\ b_3&=\binom 30 12+\binom 31 28+\binom 32 22 +\binom 33 6&&=168\\ b_4&=\binom 40 12+\binom 41 28+\binom 42 22 +\binom 43 6&&=280\\ b_5&=\binom 50 12+\binom 51 28+\binom 52 22 +\binom 53 6&&=432\\ \vdots\\ \color{red}{b_n}&\color{red}{=\binom n0 12+\binom n1 28 + \binom n2 22 +\binom n3 6}\\ &\color{red}{=n^3+8n^2+19n+12}\\ &\color{red}{=(n+1)(n+3)(n+4)} \end{align}$$

Note that $a_r$ is the first term of the $r$-th difference series, with $r=0$ referring to the original series.

The sum of the first $n$ terms is given by

$$\begin{align} S_n &=\sum_{r=0}^n \binom r0 12 +\underbrace{\sum_{r=1}^n\binom r1 28 + \underbrace{\sum_{r=2}^n\binom r2 22 +\underbrace{\sum_{r=3}^n\binom r3 6}_{n\ge 3}}_{n\ge 2}}_{n\ge 1}\\\\ &=\color{red}{\binom {n+1}1 12 +\underbrace{\binom {n+1}2 28 +\underbrace{\binom {n+1}3 22 +\underbrace{\binom {n+1}4 6}_{n\ge 3}}_{n\ge 2}}_{n\ge 1}}\\ &\color{red}{=(n+1)(3n^3+13n^2-2n+12)\qquad [\text{for }n\ge 3]} \end{align}$$


NB: References above to the $n$-th term count from $n=0$.

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