\(\left|\begin{array}{c}\mathrm{n}-1 \\\ 1\end{array}\right|+\left|\begin{array}{c}\mathrm{n}-1 \\\ 2\end{array}\right|+\ldots+\left|\begin{array}{l}\mathrm{n}-1 \\\ \mathrm{n}-1\end{array}\right|=\underline{ } ; \mathrm{n}>1\) (a) \(2^{\mathrm{n}}-1\) (b) \(2^{\mathrm{n}-2}\) (c) \(2^{\mathrm{n}-1}-1\) (d) \(2^{\mathrm{n}-1}\)

Consider the given summation: \[\left|\begin{array}{c}\mathrm{n}-1 \\\ 1\end{array}\right|+\left|\begin{array}{c}\mathrm{n}-1 \\\ 2\end{array}\right|+\ldots+\left|\begin{array}{l}\mathrm{n}-1 \\\ \mathrm{n}-1\end{array}\right|\] Notice that in each term, the absolute value expressions have the form: \[\left|\begin{array}{c}\mathrm{n}-1 \\\ \mathrm{k}\end{array}\right|\] where \(1 \leq k \leq n-1\).

Now let's individually evaluate the terms when k takes values from 1 to n-1: - When k = 1, we have: \(\left|\begin{array}{c}\mathrm{n}-1 \\\ 1\end{array}\right| = |\mathrm{n}-2|\) - When k = 2, we have: \(\left|\begin{array}{c}\mathrm{n}-1 \\\ 2\end{array}\right| = |\mathrm{n}-3|\) The pattern continues this way for all values of k. We can rewrite the initial summation as: \[|\mathrm{n}-2| + |\mathrm{n}-3| + \cdots + |\mathrm{1}| = (n-2) + (n-3) + \cdots + 1\]

Since we now have a simplified summation, it is easier to find its value. The summation is the sum of consecutive integers from 1 to n-2, which can be calculated by using the formula for the sum of an arithmetic series: Sum = \(\frac{(n-2)(n-2+1)}{2}\) Sum = \(\frac{(n-2)(n-1)}{2}\) Now we can calculate the result as follows: Sum = \(\frac{(n-2)(n-1)}{2}\) Sum = \(\frac{n^2 - 3n + 2}{2}\)

In the given options, (a), (b), (c), and (d), we need to find the one that is equivalent to the calculated summation. Simplifying each alternative: (a) \(2^{\mathrm{n}}-1\) (b) \(2^{\mathrm{n}-2}\) (c) \(2^{\mathrm{n}-1}-1\) (d) \(2^{\mathrm{n}-1}\) None of these simplified alternatives match the calculated summation exactly. However, alternative (c) comes closest to representing the summation: \(2^{\mathrm{n}-1}-1\) = \(\frac{2^{n} - 2}{2}\) = \(\frac{n^2-2n}{2}\) This is close to the calculated summation: \(\frac{n^2 - 3n + 2}{2}\) But it is not a perfect match. Thus, none of the alternatives match the calculated summation precisely. However, alternative (c) comes closest among the available options.