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 analyze the given sequence of matrices \( M_n \) and find the limit of their determinant as \( n \) approaches infinity, which is given to be equal to \( \lambda - e^{-1} \). We will derive the value of \( \lambda \) step by step.

### Step 1: Define the Matrix \( M_n \)

The matrix \( M_n \) is defined as:

\[ M_n = \begin{pmatrix} \frac{1}{(2n+1)!} & \frac{1}{(2n+2)!} \\ \sum_{k=0}^{n} \frac{(2n+2)!}{(2k+2)!} & \sum_{k=0}^{n} \frac{(2n+1)!}{(2k+1)!} \end{pmatrix} \]

### Step 2: Calculate the Determinant of \( M_n \)

The determinant of a \( 2 \times 2 \) matrix \( \begin{pmatrix} a & b \\ c & d \end{pmatrix} \) is given by \( ad - bc \).

For our matrix \( M_n \):

- \( a = \frac{1}{(2n+1)!} \)
- \( b = \frac{1}{(2n+2)!} \)
- \( c = \sum_{k=0}^{n} \frac{(2n+2)!}{(2k+2)!} \)
- \( d = \sum_{k=0}^{n} \frac{(2n+1)!}{(2k+1)!} \)

Thus, the determinant \( \text{det}(M_n) \) is:

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

### Step 3: Simplify the Determinant

The terms involving factorials can be simplified:

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

This simplifies to:

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

### Step 4: Analyze the Limit as \( n \to \infty \)

As \( n \) approaches infinity, the series \( \sum_{k=0}^{n} \frac{1}{(2k+1)!} \) approaches the Taylor series expansion for \( e^x \) evaluated at \( x = 1 \):

\[ \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} = \sinh(1) \]

The second term involving \( \frac{1}{(2n+2)!} \) will tend to zero as \( n \) approaches infinity.

Thus, we find:

\[ \lim_{n \to \infty} \text{det}(M_n) = \sum_{k=0}^{\infty} \frac{1}{(2k+1)!} = \sinh(1) \]

### Step 5: Relate to Given Limit Condition

According to the problem, we have:

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

Setting this equal to our earlier result:

\[ \sinh(1) = \lambda - e^{-1} \]

### Step 6: Solve for \( \lambda \)

Rearranging gives:

\[ \lambda = \sinh(1) + e^{-1} \]

### Conclusion

Thus, the value of \( \lambda \) is:

\[ \lambda = \sinh(1) + e^{-1} \]