The given series is: $1.2 .4+3.3 .8+5.4 .12+7.5 .16+\ldots \ldots \ldots .$

There are three series,

First series is: $1,3,5,7, \ldots$

$n^{t h}$ term of first series is:

$\begin{aligned} a_{n} &=a+(n-1) d \\ &=1+(n-1) 2 \\ &=2 n-1 \end{aligned}$

Second series is: $2,3,4,5, \ldots \ldots$

$n^{t h}$ term of second series is:

$\begin{aligned} a_{n} &=a+(n-1) d \\ &=2+(n-1) 1 \\ &=n+1 \end{aligned}$

Third series is: $4,8,12,16, \ldots \ldots$

$n^{t h}$ term of third series is:

$\begin{aligned} a_{n} &=a+(n-1) d \\ &=4+(n-1) 4 \\ &=4 n \end{aligned}$

By inspection of the given series, the $n^{t h}$ term of series is:

$\begin{aligned} T_{n} &=(2 n-1)(n+1) \times 4 n \\ &=8 n^{3}+4 n^{2}-4 n \end{aligned}$

Hence the sum of $n^{t h}$ term is:

$\begin{aligned} S_{n} &=\sum 8 n^{3}+4 n^{2}-4 n \\ &=8 \sum n^{3}+4 \sum n^{2}-4 \sum n \end{aligned}$

$=8\left(\frac{n(n+1)}{2}\right)^{2}+4 \times \frac{n(2 n+1)(n+1)}{6}-4 \times \frac{n(n+1)}{2}$

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

$=\frac{1}{2} n(n+1)\left(\frac{12 n(n+1)+4(2 n+1)-12}{3}\right)$

$=\frac{1}{2} n(n+1)\left(\frac{12 n^{2}+12 n+8 n+4-12}{3}\right)$

$=\frac{1}{2} n(n+1)\left(\frac{12 n^{2}+20 n-8}{3}\right) \\ =\frac{1}{6} n(n+1)\left(12 n^{2}+20 n-8\right)$

Sequence Introduction