If M_k= bmatrix k & k-1 k-1 & k bmatrix where k is a natural...

Given $M_k = \begin{bmatrix} k & k-1 \\ k-1 & k \end{bmatrix}$

The determinant of $M_k$ is given by:

$|M_k| = k^2 - (k-1)^2$

$|M_k| = k^2 - (k^2 - 2k + 1) = 2k - 1$

The required sum is:

$|M_1| + |M_2| + |M_3| + \ldots + |M_{50}| = \sum_{k=1}^{50} (2k - 1)$

This represents the sum of the first $50$ odd natural numbers, which forms an arithmetic progression:

$1 + 3 + 5 + \ldots + 99$

The sum of an arithmetic progression is given by $\dfrac{n}{2}(a + l)$, where $n = 50$, $a = 1$, and $l = 99$:

Sum $= \dfrac{50}{2} (1 + 99) = 25 \times 100 = 2500$

Answer: $2500$