If `L=lim_(nto oo) (n^(3)(e^(1//n)+e^(2//n)+………+e))/((n+1)^(m)(1^(m)+4^(m)+….+n^(2m)))` is non zero finite real, then

Options

A. `L=3(e-1)`
B. `L=2(e-1)`
C. `m=1//3`
D. `m=1//3`

Explanation

To solve the problem, we need to evaluate the limit

\[ L = \lim_{n \to \infty} \frac{n^3 \left( e^{1/n} + e^{2/n} + \ldots + e^{n/n} \right)}{(n+1)^m \left( 1^m + 4^m + \ldots + n^{2m} \right)} \]

### Step 1: Rewrite the numerator
The numerator can be expressed as a summation:

\[ e^{1/n} + e^{2/n} + \ldots + e^{n/n} = \sum_{r=1}^{n} e^{r/n} \]

As \( n \to \infty \), this summation can be approximated by the integral:

\[ \sum_{r=1}^{n} e^{r/n} \approx n \int_0^1 e^x \, dx \]

### Step 2: Evaluate the integral
Now we compute the integral:

\[ \int_0^1 e^x \, dx = [e^x]_0^1 = e - 1 \]

Thus, we can approximate the numerator as:

\[ \sum_{r=1}^{n} e^{r/n} \approx n(e - 1) \]

### Step 3: Rewrite the denominator
The denominator can be rewritten as:

\[ 1^m + 4^m + \ldots + n^{2m} = \sum_{r=1}^{n} (r^2)^m = \sum_{r=1}^{n} r^{2m} \]

Using the formula for the sum of powers, we have:

\[ \sum_{r=1}^{n} r^{2m} \approx \frac{n^{2m+1}}{2m+1} \]

### Step 4: Rewrite the limit
Substituting the approximations into the limit, we have:

\[ L = \lim_{n \to \infty} \frac{n^3 \cdot n(e - 1)}{(n+1)^m \cdot \frac{n^{2m+1}}{2m+1}} \]

### Step 5: Simplify the expression
This simplifies to:

\[ L = \lim_{n \to \infty} \frac{n^4(e - 1)(2m + 1)}{(n + 1)^m n^{2m + 1}} \]

As \( n \to \infty \), \((n + 1)^m \approx n^m\), so we can write:

\[ L = \lim_{n \to \infty} \frac{n^4(e - 1)(2m + 1)}{n^m n^{2m + 1}} = \lim_{n \to \infty} \frac{(e - 1)(2m + 1)}{n^{m - 4}} \]

### Step 6: Determine conditions for non-zero finite limit
For \(L\) to be non-zero and finite, we require:

\[ m - 4 = 0 \implies m = 4 \]

### Conclusion
Thus, the value of \(m\) such that \(L\) is a non-zero finite real number is:

\[ \boxed{4} \]