Explanation

(a)
Here, $T_{n}\ of\ the\ AP\ 1,2,3\ldots = n$
and${\ T}_{n\ }$of the AP $3,5,7\ldots = 2n + 1$
$\therefore T_{n}$of given series$= n(2n + 1)^{2} = 4n^{3} + 4n^{2} + n$
Hence, $S = \sum_{n = 1}^{20}{T_{n} = 4\sum_{n = 1}^{20}{n^{3} + 4\sum_{n = 1}^{20}{n^{2} + \sum_{n = 1}^{20}n}}}$
$= 4\frac{1}{4}20^{2}.21^{2} + 4\frac{1}{6}20.21.41$+$\frac{1}{2}20.21$
$= 188090$