MCQ
$\frac{1}{1} + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + ......(n + 1)$ પદ સુધી ${\text{ = }}.........$
- A$\frac{n}{{n + 1}}$
- B$\frac{{2n}}{{n + 1}}$
- C$\frac{2}{{n(n + 1)}}$
- D$\frac{{2(n + 1)}}{{n + 2}}$
${t_n} = \frac{1}{{\sum n }} = \frac{1}{{\frac{{n(n + 1)}}{2}}} = 2\left[ {\frac{1}{n} - \frac{1}{{n + 1}}} \right]$
હવે, $ n = 1,2,3,..... (n+1)$ મુક્તા
${t_1} = 2\left[ {\frac{1}{1} - \frac{1}{2}} \right],\,{t_2} = 2\left[ {\frac{1}{2} - \frac{1}{3}} \right],.....,\,\,\,{t_{n + 1}} = 2\left[ {\frac{1}{{n + 1}} - \frac{1}{{n + 2}}} \right]\,\,\,\,$
$\therefore \,\,\frac{1}{1} + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + .....(n + 1)$ પદ સુધી
$ = \sum\limits_{k = 1}^{n + 1} {{t_k}} = {S_{n + 1}} = 2\left[ {1 - \frac{1}{{n + 2}}} \right] = 2\left( {\frac{{n + 1}}{{n + 2}}} \right)$
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