MCQ
If the sum of the first $n$ terms of the series $\sqrt 3 + \sqrt {75} + \sqrt {243} + \sqrt {507} + ......$ is $435\sqrt 3 $ , then $n$ equals
- A$18$
- ✓$15$
- C$13$
- D$29$
$ \Rightarrow \sqrt 3 = \left[ {1 + 5 + 9 + 13 + ..... + {T_n}} \right] = 435\sqrt 3 $
$ \Rightarrow \sqrt 3 \times \frac{n}{2}\left[ {2 + \left( {n - 1} \right)4} \right] = 435\sqrt 3 $
$ \Rightarrow 2n + 4{n^2} - 4n = 870$
$ \Rightarrow 4{n^2} - 2n - 870 = 0$
$ \Rightarrow 2{n^2} - n - 435 = 0$
$ \Rightarrow n = \frac{{1 \pm \sqrt 1 + 4 \times 2 \times 432}}{4} = \frac{{1 \pm 59}}{4}$
$\therefore \,\,\,\,n = \frac{{1 \pm 59}}{4} = 15$; or $ = \frac{{1 - 59}}{4} = 14.5$
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$\lim _{\alpha \rightarrow 0}\left(\frac{e^{\cos \left(\alpha^n\right)}-e}{\alpha^m}\right)=-\left(\frac{e}{2}\right)$
then the value of $\frac{m}{n}$ is