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Mathematics
Mathematics
Problem
If
2
tan
2
θ
−
5
sec
θ
=
1
2\tan^2 \theta - 5\sec \theta = 1
2
tan
2
θ
−
5
sec
θ
=
1
has exactly 7 solutions in the interval
[
0
,
n
π
2
]
\left[0, \frac{n\pi}{2}\right]
[
0
,
2
nπ
]
, for the least value of
n
∈
N
n \in \mathbb{N}
n
∈
N
then
∑
k
=
1
n
k
2
k
\sum_{k=1}^{n} \frac{k}{2^k}
∑
k
=
1
n
2
k
k
is equal to: (A)
1
2
15
(
2
14
−
14
)
\frac{1}{2^{15}}(2^{14} - 14)
2
15
1
(
2
14
−
14
)
(B)
1
2
14
(
2
15
−
15
)
\frac{1}{2^{14}}(2^{15} - 15)
2
14
1
(
2
15
−
15
)
(C)
1
−
15
2
13
1 - \frac{15}{2^{13}}
1
−
2
13
15
(D)
1
2
13
(
2
14
−
15
)
\frac{1}{2^{13}}(2^{14} - 15)
2
13
1
(
2
14
−
15
)
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