A) \[\frac{13}{36}\]
B) \[\frac{12}{36}\]
C) \[\frac{15}{36}\]
D) \[\frac{18}{36}\]
Correct Answer: C
Solution :
\[{{T}_{n}}=\frac{1}{{{n}^{2}}+(n-2)}=\frac{1}{(n+2)(n-1)},\] |
\[n=3,4,5,.....\]\[=\frac{1}{3}\left[ \frac{1}{n-1}-\frac{1}{n+2} \right]\] |
\[\therefore S=\sum\limits_{n=3}^{\infty }{{{T}_{n}}=\frac{1}{3}\left( \frac{1}{2}-\frac{1}{5} \right)}+\frac{1}{3}\left( \frac{1}{3}-\frac{1}{6} \right)\]\[+\frac{1}{3}\left( \frac{1}{4}-\frac{1}{7} \right)+\frac{1}{3}\left( \frac{1}{5}-\frac{1}{8} \right)\] |
: : |
????? |
\[S=\frac{1}{3}\left[ \frac{1}{2}+\frac{1}{3}+\frac{1}{4} \right]=\frac{1}{3}\left[ \frac{6+4+3}{12} \right]=\frac{13}{36}\] |
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