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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
$\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ........ + .......\frac{1}{{n.(n + 1)}}$ equals
  • A
    $\frac{1}{{n(n + 1)}}$
  • ✓
    $\frac{n}{{n + 1}}$
  • C
    $\frac{{2n}}{{n + 1}}$
  • D
    $\frac{2}{{n(n + 1)}}$
✓

Answer

Correct option: B.
$\frac{n}{{n + 1}}$
b
(b) $\left( {\frac{1}{1} - \frac{1}{2}} \right) + \left( {\frac{1}{2} - \frac{1}{3}} \right) + \left( {\frac{1}{3} - \frac{1}{4}} \right) + ......... + \left( {\frac{1}{n} - \frac{1}{{n + 1}}} \right)$

$ = 1 - \frac{1}{{n + 1}} = \frac{n}{{n + 1}}$.

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