If `n_(1)`, `n_(2)`, `n_(3)`,…..,`n_(100)` are positive real numbers such that `n_(1)+n_(2)+n_(3)+…+n_(100)=20` and `k=n_(1)(n_(2)+n_(3)+n_(4))(n_(5)+n_(6)+…+n_(9))(n_(10)+….+n_(16))…(…+n_(100))`, then `k` belongs to

Options

A. `(o,100]`
B. `(o,128]`
C. `(o,144]`
D. `(o,1024]`

Explanation

To solve the problem, we need to analyze the expression for \( k \) and apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality.

### Step-by-Step Solution:

1. **Understanding the Problem**:
We have positive real numbers \( n_1, n_2, n_3, \ldots, n_{100} \) such that:
\[ n_1 + n_2 + n_3 + \ldots + n_{100} = 20 \]
We need to evaluate:
\[ k = n_1(n_2 + n_3 + n_4)(n_5 + n_6 + n_7 + n_8 + n_9)(n_{10} + n_{11} + \ldots + n_{16}) \ldots (n_{97} + n_{98} + n_{99} + n_{100}) \]

2. **Identifying the Groups**:
The terms in \( k \) can be grouped as follows:
- The first term is \( n_1 \).
- The second term is the sum of 3 numbers: \( n_2 + n_3 + n_4 \).
- The third term is the sum of 5 numbers: \( n_5 + n_6 + n_7 + n_8 + n_9 \).
- The fourth term is the sum of 7 numbers: \( n_{10} + n_{11} + \ldots + n_{16} \).
- Continuing this pattern, we will eventually cover all 100 terms.

3. **Finding the Number of Groups**:
The number of terms in each group follows the sequence: 1, 3, 5, 7, ..., which is an arithmetic series. The total number of terms in this series must equal 100:
\[ 1 + 3 + 5 + \ldots + (2n - 1) = n^2 \]
Setting \( n^2 = 100 \) gives \( n = 10 \). Thus, there are 10 groups.

4. **Applying AM-GM Inequality**:
By the AM-GM inequality, we know that:
\[ \frac{n_1 + (n_2 + n_3 + n_4) + (n_5 + n_6 + n_7 + n_8 + n_9) + \ldots + (n_{97} + n_{98} + n_{99} + n_{100})}{10} \geq \sqrt[10]{k} \]
The sum of all groups equals 20, so:
\[ \frac{20}{10} \geq \sqrt[10]{k} \]
Simplifying gives:
\[ 2 \geq \sqrt[10]{k} \]

5. **Raising Both Sides to the Power of 10**:
Raising both sides to the power of 10:
\[ 2^{10} \geq k \]
Calculating \( 2^{10} \):
\[ 1024 \geq k \]

6. **Finding the Range of \( k \)**:
Since \( n_i \) are positive real numbers, \( k \) must also be positive. Thus, we have:
\[ {0 < k < 1024} \]

### Conclusion:
Therefore, the final answer is:
\[ k \text{ belongs to } (0, 1024) \]