8. sequence and series — MATHS STD 11 Science — Question

and ${T_n}$ of the $A.P.$ $3,\;5,\;7,........ = 2n + 1$

$\therefore $${T_n}$ of given series $ = n{(2n + 1)^2} = 4{n^3} + 4{n^2} + n$

Hence $S = \sum\limits_1^{20} {{T_n}} = 4\sum\limits_1^{20} {{n^3}} + 4\sum\limits_1^{20} {{n^2}} + \sum\limits_1^{20} n $

$ = 4 \cdot \frac{1}{4}{20^2} \cdot {21^2} + 4 \cdot \frac{1}{6}20 \cdot 21 \cdot 41 + \frac{1}{2}20 \cdot 21 = 188090$.