summation - Alternative ways to find the sum of this series

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 $=S_k=\frac{k}{2}[2\times 4+(k-1)2 ]=\frac{k}{2}[8+2k-2]=\frac{k}{2}(2k+6)=k(k+3)$

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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?