1.2.4 + 2.3.7 +3.4.10 + ...

Let $T_{n}$ be the $n$th term of the given series.

Thus, we have:

$T_{n}=n(n+1)(3 n+1)=n\left(3 n^{2}+4 n+1\right)=\left(3 n^{3}+4 n^{2}+n\right)$

Now, let $S_{n}$ be the sum of $n$ terms of the given series.

Thus, we have:

$S_{n}=\sum_{k=1}^{n} T_{k}$

$\Rightarrow S_{n}=\sum_{k=1}^{n} 3 k^{3}+\sum_{k=1}^{n} 4 k^{2}+\sum_{k=1}^{n} k$

$\Rightarrow S_{n}=3 \sum_{k=1}^{n} k^{3}+4 \sum_{k=1}^{n} k^{2}+\sum_{k=1}^{n} k$

$\Rightarrow S_{n}=\frac{3 n^{2}(n+1)^{2}}{4}+\frac{4 n(n+1)(2 n+1)}{6}+\frac{n(n+1)}{2}$

$\Rightarrow S_{n}=\frac{3 n^{2}(n+1)^{2}}{4}+\frac{2 n(n+1)(2 n+1)}{3}+\frac{n(n+1)}{2}$

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

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{3 n^{2}+3 n}{2}+\frac{8 n+4}{3}+1\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{2}\left(\frac{9 n^{2}+9 n+16 n+8+6}{6}\right)$

$\Rightarrow S_{n}=\frac{n(n+1)}{12}\left(9 n^{2}+25 n+14\right)$