If 2 ^ 2 θ - 5secθ = 1 has exactly 7 solutions in the interval [0,...

Given,

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

$⇒ 2\left(\sec^{2}\theta - 1\right) - 5sec \theta - 1 = 0$

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

$⇒ 2\sec^{2}\theta - 6sec \theta + sec \theta - 3 = 0$

$⇒ 2sec \theta\left(sec\theta - 3\right) + 1\left(sec \theta - 3\right) = 0$

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

$⇒ sec \theta = - \dfrac{1}{2}, 3$

$⇒ cos \theta = - 2, \dfrac{1}{3}$

We know that, $- 1 \le cos \theta \le 1$

$⇒ cos \theta = \dfrac{1}{3}$

For $7$ solutions $n = 13$

So, $\mathop{∑}_{k = 1}^{13}\dfrac{k}{{2}^{k}} = S$ (say)

$⇒ S = \dfrac{1}{2} + \dfrac{2}{{2}^{2}} + \dfrac{3}{{2}^{3}} + \ldots . + \dfrac{13}{{2}^{13}}$

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

$⇒ S - \dfrac{S}{2} = \left(\dfrac{1}{2} + \dfrac{2}{{2}^{2}} + \dfrac{3}{{2}^{3}} + \ldots . + \dfrac{13}{{2}^{13}}\right) - \left(\dfrac{1}{{2}^{2}} + \dfrac{1}{{2}^{3}} + \ldots . . + \dfrac{12}{{2}^{13}} + \dfrac{13}{{2}^{14}}\right)$

$⇒ \dfrac{S}{2} = \left(\dfrac{1}{2} + \dfrac{1}{{2}^{2}} + \dfrac{3}{{2}^{3}} + . . . + \dfrac{13}{{2}^{13}}\right) - \dfrac{13}{{2}^{14}}$

$⇒ \dfrac{S}{2} = \dfrac{1}{2} \cdot \dfrac{1 - \dfrac{1}{{2}^{13}}}{1 - \dfrac{1}{2}} - \dfrac{13}{{2}^{14}}$

$⇒ \dfrac{S}{2} = \dfrac{1}{2} \cdot \dfrac{\dfrac{{2}^{13} - 1}{{2}^{13}}}{\dfrac{1}{2}} - \dfrac{13}{{2}^{14}}$

$⇒ S = 2 \cdot \left(\dfrac{{2}^{13} - 1}{{2}^{13}}\right) - \dfrac{13}{{2}^{13}}$

$⇒ S = \left(\dfrac{{2}^{14} - 2 - 13}{{2}^{13}}\right)$

$⇒ S = \left(\dfrac{{2}^{14} - 15}{{2}^{13}}\right)$