Given that
`lim_(n to oo)sum_(r=1)^(n)(log(n^(2)+r^(2))-2logn)/(n)=log2+(pi)/(2)-2`, then
`lim_(n to oo) (1)/(n^(2m))[(n^(2)+1^(2))^(m)(n^(2)+r^(2))^(m)......(n^(2))^(m)]^(1//n)` is equal to

Options

A. `2^(m)e^(m((pi)/(2)-2))`
B. `2^(m)e^(m(2-(pi)/(2)))`
C. `e^m((pi)/(2)-2)`
D. `e^(2m((pi)/(2)-2))`

Answer

To solve the given problem step by step, we will analyze the limit and the summation provided.

### Step 1: Understanding the Given Limit
We start with the limit:
\[ \lim_{n \to \infty} \frac{1}{n^{2m}} \left[(n^2 + 1^2)^m (n^2 + r^2)^m \cdots (n^2)^m\right]^{\frac{1}{n}} \]
This expression involves a product of terms raised to the power \( m \) and then taken to the \( \frac{1}{n} \) power.

### Step 2: Rewriting the Expression
We can rewrite the expression inside the limit:
\[ \lim_{n \to \infty} \frac{1}{n^{2m}} \left[(n^2 + 1^2)(n^2 + 2^2) \cdots (n^2 + n^2)\right]^m \]
This means we need to analyze the product:
\[ (n^2 + 1^2)(n^2 + 2^2) \cdots (n^2 + n^2) \]

### Step 3: Factor Out \( n^2 \)
Notice that each term can be factored as:
\[ n^2 + r^2 = n^2(1 + \frac{r^2}{n^2}) \]
Thus, we can express the product as:
\[ \prod_{r=1}^{n} (n^2 + r^2) = n^{2n} \prod_{r=1}^{n} \left(1 + \frac{r^2}{n^2}\right) \]

### Step 4: Substitute Back into the Limit
Substituting this back into our limit gives:
\[ \lim_{n \to \infty} \frac{1}{n^{2m}} \left[n^{2n} \prod_{r=1}^{n} \left(1 + \frac{r^2}{n^2}\right)\right]^m \]
This simplifies to:
\[ \lim_{n \to \infty} n^{2n \cdot m - 2m} \left(\prod_{r=1}^{n} \left(1 + \frac{r^2}{n^2}\right)\right)^m \]

### Step 5: Analyze the Product
The product \( \prod_{r=1}^{n} \left(1 + \frac{r^2}{n^2}\right) \) can be approximated using the logarithm:
\[ \log \left(\prod_{r=1}^{n} \left(1 + \frac{r^2}{n^2}\right)\right) = \sum_{r=1}^{n} \log\left(1 + \frac{r^2}{n^2}\right) \]
As \( n \to \infty \), this sum can be approximated by an integral:
\[ \sum_{r=1}^{n} \log\left(1 + \frac{r^2}{n^2}\right) \sim n \int_0^1 \log(1 + x^2) \, dx \]

### Step 6: Final Limit Calculation
Thus, we have:
\[ \lim_{n \to \infty} \frac{1}{n^{2m}} n^{2n \cdot m - 2m} e^{m \cdot n \int_0^1 \log(1 + x^2) \, dx} \]
The limit simplifies to:
\[ e^{m \cdot n \int_0^1 \log(1 + x^2) \, dx} \text{ as } n \to \infty \]

### Step 7: Conclusion
The final result is:
\[ 2^m e^{m \cdot \left(\frac{\pi}{2} - 2\right)} \]