Q99 / 180 · KCET 2008 · Mathematics

$\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+$ equal to upto $n$ terms is

  • A
    $\frac{n}{4 n+6}$
  • B
    $\frac{1}{6 n+4}$
  • Correct answer
    C
    $\frac{n}{6 n+4}$
  • D
    $\frac{n}{3 n+7}$

Solution

Let $$ \begin{aligned} \mathrm{S}_{\mathrm{n}} &=\frac{1}{2 \cdot 5}+\frac{1}{5 \cdot 8}+\frac{1}{8 \cdot 11}+\ldots+\frac{1}{(3 n-1)(3 n+2)} \\ &=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{5}+\frac{1}{5}-\frac{1}{8}+\ldots \ldots+\frac{1}{3 n-1}-\frac{1}{3 n+2}\right] \\ &=\frac{1}{3}\left[\frac{1}{2}-\frac{1}{3 n+2}\right] \\ &=\frac{n}{6 n+4} \end{aligned} $$

Hence, the correct answer is option: 'C'.

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