Question: Find the sum of the series $3+7+13+21+\dotsm$ upto $n$ terms.
My attempt:
Consider the arithmetic series $4+6+8+\dotsm$ upto $n$ terms.
The sum of first $k$ terms of this series $=S_k=\frac{k}{2}[2\times 4+(k-1)2 ]=\frac{k}{2}[8+2k-2]=\frac{k}{2}(2k+6)=k(k+3)$
Now, consider the sequence $3,7,13,21,\dotsm$ upto $n$ terms.
This sequence can be written as $3,(3+4),(3+4+6),(3+4+6+8),\dotsm$ upto $n$ terms.
General term of this sequence $=t_k=3+S_{k-1}=3+(k-1)(k+2)=3+k^2+2k-k-2=k^2+k+1$
Therefore the sum of the series $3+7+13+21+\dotsm$ upto $n$ terms $=S_n^{'}=\sum_{k=1}^{n}t_k=\sum_{k=1}^{n}(k^2+k+1)=\sum_{k=1}^{n}k^2+\sum_{k=1}^{n}k+\sum_{k=1}^{n}1=\frac{n(n+1)(2n+1)}{6}+\frac{n(n+1)}{2}+n=n[\frac{(n+1)(2n+1)}{6}+\frac{n+1}{2}+1]=n(\frac{2n^2+3n+1+3n+3+6}{6})=\frac{n}{6}(2n^2+6n+10)=\frac{n}{3}(n^2+3n+5)$
My problem: Is there any elegant method to find the sum of this series?
The formula for the sum of a general quadratic sequence is:
$$S_2(n,s,d_1,c) = \cfrac{n(cn^2 + 2c + 3nd_1 + 6s - 3cn - 3d_1)}{6}$$
Where $n$ is the number of terms to be summed, $s$ is the starting term of the series, $d_1$ is the first difference (subtracting the first term from the second term) and $c$ is the constant difference between the differences.
From your series we can see that $s = 3$, $d_1 = 4$, $c = 2$.
Plug in the values in the sum function:
$$S_2(n,3,4,2) = \cfrac{n(2n^2 + 4 + 12n + 18 - 6n - 12)}{6}$$ Simplify: $$ S_2(n,3,4,2) = \frac{n^3 + 5n + 3n^2}{3}$$
I have the proof of this sum function somehere on a document, if you want it I can dig it up. It's rather old and messy though.