MCQ
If ${S_n} = \frac{{n(n + 1)\left( {n + 2} \right)}}{6}$ then $\sum\limits_{n = 1}^\infty {\frac{1}{{{t_n}}}} = $
- A$1$
- B$6$
- ✓$2$
- D$\frac {1}{6}$
$\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)(\mathrm{n}+2)}{6}-\frac{(\mathrm{n}-1) \mathrm{n}(\mathrm{n}+1)}{6}$
$\mathrm{t}_{\mathrm{n}}=\frac{\mathrm{n}(\mathrm{n}+1)}{6}[\mathrm{n}+2-\mathrm{n}+1]=\frac{\mathrm{n}(\mathrm{n}+1)}{2}$
$\sum\limits_{n = 1}^\infty {\frac{2}{{n\left( {n + 1} \right)}}} = 2\left[ {\frac{1}{1} - \frac{1}{2} + \frac{1}{2} - \frac{1}{3}.......} \right] = 2$
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