JEE Main 2024 — 27 January, Shift 2 — Question 15

$2 \tan ^{2} \theta-5 \sec \theta-1=0$

$\Rightarrow 2 \sec ^{2} \theta-5 \sec \theta-3=0$

$\Rightarrow(2 \sec \theta+1)(\sec \theta-3)=0$

$\Rightarrow \sec \theta=-\frac{1}{2}, 3$

$\Rightarrow \cos \theta=-2, \frac{1}{3}$

$\Rightarrow \cos \theta=\frac{1}{3}$

For 7 solutions $\mathrm{n}=13$

So, $\sum_{\mathrm{k}=1}^{13} \frac{\mathrm{k}}{2^{\mathrm{k}}}=\mathrm{S}$ (say) $\mathrm{S}=\frac{1}{2}+\frac{2}{2^{2}}+\frac{3}{2^{3}}+\ldots .+\frac{13}{2^{13}}$

$\frac{1}{2} S=\frac{1}{2^{2}}+\frac{1}{2^{3}}+\ldots . .+\frac{12}{2^{13}}+\frac{13}{2^{14}}$

$\Rightarrow \frac{S}{2}=\frac{1}{2} \cdot \frac{1-\frac{1}{2^{13}}}{1-\frac{1}{2}}-\frac{13}{2^{14}} \Rightarrow S=2 .\left(\frac{2^{13}-1}{2^{13}}\right)-\frac{13}{2^{13}}$

$$\sum\limits_{k = 1}^n {\frac{k}{{{2^k}}}} =\frac{1}{2^{13}}\left(2^{14}-15\right)$$