STD 11 - 8. sequence and series — JEE STD 11 Science — Question

${n^{th}}$ term ${T_n} = {\left( {2n - 1} \right)^2},n = 1,.....13$

Now, ${S_n} = \,\sum\limits_{n = 1}^{13} {{T_n} = } \sum\limits_{n = 1}^{13} {{{\left( {2n - 1} \right)}^2}} $

$ = \sum\limits_{n = 1}^{13} {4{n^2} + \sum\limits_{n = 1}^{13} {1 - \sum\limits_{n = 1}^{13} {4n} } } $

$ = 4\sum {{n^2} + 13 - 4\sum n } $

$ = 4\left[ {\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}} \right] + 13 - 4\frac{{n\left( {n + 1} \right)}}{2}$

Put $n=13$, we get

${S_n} = 26 \times 14 \times 9 + 13 - 26 \times 14$

$ = 3276 + 13 - 364$

$ = 2925$