If K = left|begin{array}{ll}3 & 4 5 & 4end{a | vedclass

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(B) Let the $n$-th term of the series be $D_n = \left|\begin{array}{cc} a_n & b_n \ 5 & 4 \end{array}ight| = 4a_n - 5b_n$. Here, $a_n$ is a geometr

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$2$

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(B) Let the $n$-th term of the series be $D_n = \left|\begin{array}{cc} a_n & b_n \ 5 & 4 \end{array}ight| = 4a_n - 5b_n$. Here, $a_n$ is a geometric progression with first term $a_1 = 3$ and common ratio $r_1 = 1/3$. Thus, $a_n = 3(1/3)^{n-1}$. Also, $b_n$ is a geometric progression with first term $b_1 = 4$ and common ratio $r_2 = -1/4$. Thus, $b_n = 4(-1/4)^{n-1}$. Therefore, $K = \sum_{n=1}^{\infty} (4a_n - 5b_n) = 4 \sum_{n=1}^{\infty} a_n - 5 \sum_{n=1}^{\infty} b_n$. Using the sum of an infinite geometric series formula $S = \frac{a}{1-r}$: $\sum_{n=1}^{\infty} a_n = \frac{3}{1 - 1/3} = \frac{3}{2/3} = \frac{9}{2}$. $\sum_{n=1}^{\infty} b_n = \frac{4}{1 - (-1/4)} = \frac{4}{5/4} = \frac{16}{5}$. Substituting these values into the expression for $K$: $K = 4 \left(\frac{9}{2}ight) - 5 \left(\frac{16}{5}ight) = 18 - 16 = 2$.

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