TS EAMCET · Maths · Determinants
\(\mathrm{K}=\left|\begin{array}{ll}3 & 4 \\ 5 & 4\end{array}\right|+\left|\begin{array}{cc}1 & -1 \\ 5 & 4\end{array}\right|+\left|\begin{array}{cc}\frac{1}{3} & \frac{1}{4} \\ 5 & 4\end{array}\right|+\left|\begin{array}{cc}\frac{1}{9} & -\frac{1}{16} \\ 5 & 4\end{array}\right|+\ldots\) to \(\infty\) then \(\mathrm{K}=\)
- A \(1\)
- B \(2\)
- C \(3\)
- D \(4\)
Answer & Solution
Correct Answer
(B) \(2\)
Step-by-step Solution
Detailed explanation
\( \mathrm{K} = \sum_{n=1}^{\infty} \left|\begin{array}{cc}a_n & b_n \\ 5 & 4\end{array}\right| = \sum_{n=1}^{\infty} (4a_n - 5b_n) = 4\sum a_n - 5\sum b_n \) \( \sum a_n = 3+1+\frac{1}{3}+\ldots = \frac{3}{1 - \frac{1}{3}} = \frac{9}{2} \)…
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