If $2 \tan ^2 \theta-5 \sec \theta=1$ has exactly 7 solutions in the interval $\left[0, \frac{n \pi}{2}\right]$, for the least value of $n \in \mathbf{N}$, then $\sum_\limits{k=1}^n \frac{k}{2^k}$ is equal to:

Question

$\frac{1}{2^{14}}\left(2^{15}-15\right)$

$1-\frac{15}{2^{13}}$

$\frac{1}{2^{15}}\left(2^{14}-14\right)$

$\frac{1}{2^{13}}\left(2^{14}-15\right)$

Solution

Please login to view the detailed solution steps...

Go to DASH