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

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 11 - 8. sequence and series4 Marks
MCQ
The value of $\sum\limits_{n = 2}^\infty  {\frac{n}{{1 + {n^2}\left( {{n^2} - 2} \right)}}} $ is equal to
  • A
    $\frac {5}{4}$
  • B
    $1$
  • ✓
    $\frac {5}{16}$
  • D
    $\frac {1}{4}$
✓

Answer

Correct option: C.
$\frac {5}{16}$
c
$\sum\limits_{n = 2}^\infty  {\frac{n}{{1 + {n^4} - 2{n^2}}}}  = \frac{1}{4}\sum\limits_{n = 2}^\infty  {\frac{{{{(n + 1)}^2} - {{(n - 1)}^2}}}{{{{(n + 1)}^2} \times {{(n - 1)}^2}}}} $

$ = \frac{1}{4}\sum\limits_{n = 2}^\infty  {\left( {\frac{1}{{{{(n - 1)}^2}}} - \frac{1}{{{{(n + 1)}^2}}}} \right) = \frac{5}{{16}}} $

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