KVPY Sample Paper KVPY Stream-SX Model Paper-16

The value of the sum \[\frac{1}{{{3}^{2}}+1}+\frac{1}{{{4}^{2}}+2}+\frac{1}{{{5}^{2}}+3}+\frac{1}{{{6}^{2}}+4}+......\infty ,\] equal to -

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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KVPY Sample Paper KVPY Stream-SX Model Paper-16

question_answer The value of the sum \[\frac{1}{{{3}^{2}}+1}+\frac{1}{{{4}^{2}}+2}+\frac{1}{{{5}^{2}}+3}+\frac{1}{{{6}^{2}}+4}+......\infty ,\] equal to - A) \[\frac{13}{36}\] B) \[\frac{12}{36}\] C) \[\frac{15}{36}\] D) \[\frac{18}{36}\]

The value of the sum \[\frac{1}{{{3}^{2}}+1}+\frac{1}{{{4}^{2}}+2}+\frac{1}{{{5}^{2}}+3}+\frac{1}{{{6}^{2}}+4}+......\infty ,\] equal to -

A) \[\frac{13}{36}\]

B) \[\frac{12}{36}\]

C) \[\frac{15}{36}\]

D) \[\frac{18}{36}\]