Evaluating Binomial Sum Given Zero Determinant Condition

To find the value of the given expression, we first evaluate each of the four binomial sums present in the entries of the matrix:

1. Top-Left Entry ($a_{11}$): $a_{11} = \sum_{k=0}^{n} k = \frac{n(n+1)}{2}$

2. Bottom-Left Entry ($a_{21}$): $a_{21} = \sum_{k=0}^{n} {}^n C_k k = \sum_{k=1}^{n} n \cdot {}^{n-1} C_{k-1} = n \sum_{j=0}^{n-1} {}^{n-1} C_j = n \cdot 2^{n-1}$

3. Top-Right Entry ($a_{12}$): $a_{12} = \sum_{k=0}^{n} {}^n C_k k^2 = \sum_{k=1}^{n} {}^n C_k (k(k-1) + k) = \sum_{k=2}^{n} n(n-1) {}^{n-2} C_{k-2} + \sum_{k=1}^{n} n {}^{n-1} C_{k-1}$ $a_{12} = n(n-1) \cdot 2^{n-2} + n \cdot 2^{n-1} = n \cdot 2^{n-2} (n - 1 + 2) = n(n+1) 2^{n-2}$

4. Bottom-Right Entry ($a_{22}$): $a_{22} = \sum_{k=0}^{n} {}^n C_k 3^k = (1 + 3)^n = 4^n = 2^{2n}$

Now, substitute these evaluated entries back into the determinant:

$$\det \begin{bmatrix} \frac{n(n+1)}{2} & n(n+1) 2^{n-2} \\ n \cdot 2^{n-1} & 2^{2n} \end{bmatrix} = 0$$

Expanding the determinant: $\left( \frac{n(n+1)}{2} \right) \cdot 2^{2n} - \left( n(n+1) 2^{n-2} \right) \cdot \left( n \cdot 2^{n-1} \right) = 0$ $\frac{n(n+1)}{2} \cdot 2^{2n} - n^2(n+1) \cdot 2^{2n-3} = 0$

Since $n$ is a positive integer ($n \ge 1$), we can divide the entire equation by $n(n+1) 2^{2n-3}$: $\frac{2^{2n-1}}{2^{2n-3}} - n = 0$ $2^2 - n = 0 \implies n = 4$

Now, we need to calculate the value of: $\sum_{k=0}^{n} \frac{{}^n C_k}{k+1}$

Using the identity $\frac{{}^n C_k}{k+1} = \frac{{}^{n+1} C_{k+1}}{n+1}$: $\sum_{k=0}^{n} \frac{{}^n C_k}{k+1} = \frac{1}{n+1} \sum_{k=0}^{n} {}^{n+1} C_{k+1} = \frac{2^{n+1} - 1}{n+1}$

For $n = 4$: $\sum_{k=0}^{4} \frac{{}^4 C_k}{k+1} = \frac{2^{4+1} - 1}{4+1} = \frac{32 - 1}{5} = \frac{31}{5} = 6.2$