If `A=[a_(ij)]_(nxxn)`, where `a_(ij)=i^100+j^100`, then `lim_(ntooo) ((overset(n)overset(suma_(ij))(i=1))/n^101)` equals

Options

A. `1/50`
B. `1/101`
C. `2/101`
D. `3/101`

Explanation

To solve the problem, we need to evaluate the limit:

\[ \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij}}{n^{101}} \]

where \( a_{ij} = i^{100} + j^{100} \).

### Step-by-step Solution:

1. **Write the expression for \( a_{ij} \)**:
\[ a_{ij} = i^{100} + j^{100} \]

2. **Calculate the double summation**:
We can express the double summation as:
\[ \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} = \sum_{i=1}^{n} \sum_{j=1}^{n} (i^{100} + j^{100}) \]
This can be split into two separate sums:
\[ = \sum_{i=1}^{n} \sum_{j=1}^{n} i^{100} + \sum_{i=1}^{n} \sum_{j=1}^{n} j^{100} \]
Since the inner summation over \( j \) for each fixed \( i \) gives \( n \cdot i^{100} \) (as \( j \) runs from 1 to \( n \)), we have:
\[ = n \sum_{i=1}^{n} i^{100} + n \sum_{j=1}^{n} j^{100} = 2n \sum_{k=1}^{n} k^{100} \]

3. **Use the formula for the sum of powers**:
The sum \( \sum_{k=1}^{n} k^{100} \) can be approximated using the formula:
\[ \sum_{k=1}^{n} k^{p} \sim \frac{n^{p+1}}{p+1} \quad \text{as } n \to \infty \]
For \( p = 100 \):
\[ \sum_{k=1}^{n} k^{100} \sim \frac{n^{101}}{101} \]

4. **Substitute back into the summation**:
Now substituting this back, we get:
\[ 2n \sum_{k=1}^{n} k^{100} \sim 2n \cdot \frac{n^{101}}{101} = \frac{2n^{102}}{101} \]

5. **Evaluate the limit**:
Now we can substitute this into our limit:
\[ \lim_{n \to \infty} \frac{\frac{2n^{102}}{101}}{n^{101}} = \lim_{n \to \infty} \frac{2n^{102}}{101n^{101}} = \lim_{n \to \infty} \frac{2n}{101} = \frac{2}{101} \]

Thus, the final result is:
\[ \lim_{n \to \infty} \frac{\sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij}}{n^{101}} = \frac{2}{101} \]

### Final Answer:
\[ \frac{2}{101} \]