If 2tan2θ - 5secθ = 1 has exactly 7 solutions in the interval \(\left[0,\frac{\mathrm{n\pi}}{2}\right]\)  , for the least value of n ϵ N then  \(\sum_{k=1}^n\frac{k}{2^k}\) is equal to :  

1
\( \frac{1}{2^{15}}\left(2^{14}-14\right)\)
2
\( \frac{1}{2^{14}}\left(2^{15}-15\right)\)
3
\( 1-\frac{15}{2^{13}}\)
4
\(\frac{1}{2^{13}}\left(2^{14}-15\right) \)

Sponsored

hivanix.in

Visit

This quiz is brought to you by hivanix.in

🌐 Web App Development

Quick Navigation