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 find the range of \( k \) defined as:

\[ k = n_1(n_2 + n_3 + n_4)(n_5 + n_6 + n_7 + n_8 + n_9)(n_{10} + n_{11} + \ldots + n_{16}) \ldots (n_{17} + n_{18} + \ldots + n_{100}) \]

given that

\[ n_1 + n_2 + n_3 + \ldots + n_{100} = 20 \]

### Step 1: Apply the AM-GM Inequality

We will use the Arithmetic Mean-Geometric Mean (AM-GM) inequality to find a lower bound for \( k \).

The AM-GM inequality states that for any non-negative real numbers \( a_1, a_2, \ldots, a_n \):

\[ \frac{a_1 + a_2 + \ldots + a_n}{n} \geq \sqrt[n]{a_1 a_2 \ldots a_n} \]

### Step 2: Group the Terms

We can group the terms in \( k \) as follows:

1. \( n_1 \)
2. \( n_2 + n_3 + n_4 \) (3 terms)
3. \( n_5 + n_6 + n_7 + n_8 + n_9 \) (5 terms)
4. \( n_{10} + n_{11} + \ldots + n_{16} \) (7 terms)
5. \( n_{17} + n_{18} + \ldots + n_{100} \) (84 terms)

### Step 3: Apply AM-GM to Each Group

Using AM-GM for each group:

1. For \( n_1 \):
\[ n_1 \geq n_1 \]

2. For \( n_2 + n_3 + n_4 \) (3 terms):
\[ \frac{n_2 + n_3 + n_4}{3} \geq \sqrt[3]{n_2 n_3 n_4} \]

3. For \( n_5 + n_6 + n_7 + n_8 + n_9 \) (5 terms):
\[ \frac{n_5 + n_6 + n_7 + n_8 + n_9}{5} \geq \sqrt[5]{n_5 n_6 n_7 n_8 n_9} \]

4. For \( n_{10} + n_{11} + \ldots + n_{16} \) (7 terms):
\[ \frac{n_{10} + n_{11} + \ldots + n_{16}}{7} \geq \sqrt[7]{n_{10} n_{11} \ldots n_{16}} \]

5. For \( n_{17} + n_{18} + \ldots + n_{100} \) (84 terms):
\[ \frac{n_{17} + n_{18} + \ldots + n_{100}}{84} \geq \sqrt[84]{n_{17} n_{18} \ldots n_{100}} \]

### Step 4: Combine the Inequalities

Using the total sum \( n_1 + n_2 + n_3 + \ldots + n_{100} = 20 \), we can write:

\[ \frac{20}{10} \geq \sqrt[10]{k} \]

This simplifies to:

\[ 2 \geq \sqrt[10]{k} \]

### Step 5: Raise Both Sides to the Power of 10

Raising both sides to the power of 10 gives:

\[ 2^{10} \geq k \]

Calculating \( 2^{10} \):

\[ 1024 \geq k \]

### Step 6: Establish the Range of \( k \)

Since \( k \) is a product of positive real numbers, we also have \( k > 0 \). Thus, we can conclude:

\[ 0 < k \leq 1024 \]

### Final Answer

The range of \( k \) is:

\[ k \in (0, 1024] \]