If 2tan2θ-5secθ=1 has exactly 7 solutions in the interval 0,nπ2, for the least value of n∈N then ∑k=1nk2k is equal to :

Question

2 \tan^{2} θ - 5 \sec θ = 1

has exactly

7

solutions in the interval

[0, \frac{n π}{2}]

, for the least value of

n \in N

then

\sum_{k = 1}^{n} \frac{k}{2^{k}}

is equal to :

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

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

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

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

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