A sequence of `2xx2` matrices `{M_(n)}` is defined as follows `M_(n)=[((1)/((2n+1)!),(1)/((2n+2)!)),(sum_(k=0)^(n)((2n+2)!)/((2k+2)!), sum_(k=0)^(n)((2n+1)!)/((2k+1)!))]` then `lim_(n to oo) "det". (M_(n))=lambda-e^(-1)`. Find `lambda`.

Answer

To solve the problem, we need to find the limit of the determinant of the matrix \( M_n \) as \( n \) approaches infinity, and then determine the value of \( \lambda \) such that:

\[ \lim_{n \to \infty} \text{det}(M_n) = \lambda - e^{-1} \]

### Step-by-Step Solution:

1. **Define the Matrix \( M_n \)**:
The matrix \( M_n \) is given by:
\[ M_n = \begin{pmatrix} \frac{1}{(2n+1)!} & \frac{1}{(2n+2)!} \\ \sum_{k=0}^{n} \frac{(2n+2)!}{(2k+2)!} & \sum_{k=0}^{n} \frac{(2n+1)!}{(2k+1)!} \end{pmatrix} \]

2. **Calculate the Determinant**:
The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is calculated as:
\[ \text{det}(M_n) = ad - bc \]
For our matrix, we have:
- \( a = \frac{1}{(2n+1)!} \)
- \( b = \frac{1}{(2n+2)!} \)
- \( c = \sum_{k=0}^{n} \frac{(2n+2)!}{(2k+2)!} \)
- \( d = \sum_{k=0}^{n} \frac{(2n+1)!}{(2k+1)!} \)

Thus,
\[ \text{det}(M_n) = \frac{1}{(2n+1)!} \cdot \sum_{k=0}^{n} \frac{(2n+1)!}{(2k+1)!} - \frac{1}{(2n+2)!} \cdot \sum_{k=0}^{n} \frac{(2n+2)!}{(2k+2)!} \]

3. **Simplify the Determinant**:
We can factor out \( \frac{(2n+1)!}{(2n+2)!} \) and \( \frac{(2n+2)!}{(2n+1)!} \):
\[ \text{det}(M_n) = \frac{1}{(2n+1)!} \left( \sum_{k=0}^{n} \frac{(2n+1)!}{(2k+1)!} - \frac{(2n+2)!}{(2n+1)!} \sum_{k=0}^{n} \frac{(2n+2)!}{(2k+2)!} \right) \]

4. **Evaluate the Limit**:
As \( n \to \infty \), the sums can be recognized as series that converge to known values:
\[ \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} = \sinh(1) \quad \text{and} \quad \sum_{k=0}^{\infty} \frac{1}{(2k+2)!} = \frac{1}{2} \sinh(1) \]
Therefore, we can evaluate:
\[ \lim_{n \to \infty} \text{det}(M_n) = 1 - e^{-1} \]

5. **Set Up the Equation**:
From the problem statement, we have:
\[ \lim_{n \to \infty} \text{det}(M_n) = \lambda - e^{-1} \]
Setting the two expressions equal gives:
\[ 1 - e^{-1} = \lambda - e^{-1} \]

6. **Solve for \( \lambda \)**:
Rearranging the equation yields:
\[ \lambda = 1 \]

### Final Answer:
Thus, the value of \( \lambda \) is:
\[ \boxed{1} \]