$\frac{1}{1 \times 5} + \frac{1}{5 \times 9} + \frac{1}{9 \times 13} + \ldots$ $n$ પદો સુધી $=$

  • A
    $\frac{1}{4n+1}$
  • B
    $\frac{4}{4n+1}$
  • C
    $\frac{n}{4n+1}$
  • D
    $\frac{4n+1}{5(4n+1)}$

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જો ${a_k} = \frac{1}{{k(k + 1)}}$ હોય,જ્યાં $k = 1, 2, 3, 4, ..., n$,તો ${\left( {\sum\limits_{k = 1}^n {{a_k}} } \right)^2} = $

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