Question

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 \( n \in \mathbb{N} \) then \(\sum_{k=1}^{n} \frac{k}{2^k}\) is equal to:

• A. \( \frac{1}{2^{15}} (2^{14} - 14) \)
• B. \( \frac{1}{2^{13}} (2^{14} - 15) \)
• C. \( \frac{1}{2^{14}} (2^{15} - 15) \)
• D. \( 1 - \frac{15}{2^{13}} \)

Answer 1

The Correct Option is B

Given: 
\(2\tan^2\theta - 5\sec\theta - 1 = 0.\)

Rewriting:
\(2\sec^2\theta - 5\sec\theta - 3 = 0.\)

Factoring:
\((2\sec\theta + 1)(\sec\theta - 3) = 0.\)

Thus:
\(\sec\theta = -\frac{1}{2}, \quad \sec\theta = 3.\)

Since \(\sec\theta = \frac{1}{\cos\theta}\), we find:  
\(\cos\theta = -2, \quad \cos\theta = \frac{1}{3}.\)

However, \(\cos\theta = -2\) is not possible, so:  
\(\cos\theta = \frac{1}{3}.\)

For the series sum:
Given:
\(S = \sum_{k=1}^{13} \frac{k}{2^k}.\)

The sum can be written as:  
\(S = \frac{1}{2} + \frac{2}{2^2} + \frac{3}{2^3} + \dots + \frac{13}{2^{13}}.\)

Multiplying the entire series by \(\frac{1}{2}\):
\(\frac{S}{2} = \frac{1}{2^2} + \frac{2}{2^3} + \dots + \frac{12}{2^{13}} + \frac{13}{2^{14}}.\)

Subtracting:
\(S - \frac{S}{2} = \frac{1}{2} + \frac{1}{2^2} + \frac{1}{2^3} + \dots + \frac{1}{2^{13}} - \frac{13}{2^{14}}.\)

Simplifying:
\(\frac{S}{2} = \left(1 - \frac{1}{2^{13}}\right) - \frac{13}{2^{14}}.\)

Thus
\(S = 2\left(1 - \frac{1}{2^{13}}\right) - \frac{13}{2^{13}}.\)

Simplifying further:
\(S = \frac{1}{2^{13}}(2^{14} - 15).\)

The correct option is (B) : \( \frac{1}{2^{13}} (2^{14} - 15) \)