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

$\frac{3}{{12}} + \frac{5}{{{1^2} + {2^2}}} + \frac{7}{{{1^2} + {2^2} + {3^2}}} + .....$

${n^{th}}$ term $ = {T_n}$

$ = \frac{{2n + 1}}{{\frac{{n\left( {n + 1} \right)\left( {2n + 1} \right)}}{6}}} = \frac{6}{{n\left( {n + 1} \right)}}$

or ${T_n} = 6\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right]$

$\therefore {S_n} = $

$\sum {{T_n} = 6\sum {\frac{1}{n} - 6} \sum {\frac{1}{{n + 1}}} }  = \frac{{6n}}{n} - \frac{6}{{n + 1}}$

$ = 6 - \frac{6}{{n + 1}} = \frac{{6n}}{{n + 1}}$

So, sum upto $11$ terms means

${S_{11}} = \frac{{6 \times 11}}{{11 + 1}} = \frac{{66}}{{12}} = \frac{{33}}{6} = \frac{{11}}{2}$