Statement-1: `1^(3)+3^(3)+5^(3)+7^(3)+...+(2n-1)^(3)ltn^(4),n in N`
Statement-2: If `a_(1),a_(2),a_(3),…,a_(n)` are n distinct positive real numbers and `mgt1`, then
`(a_(1)^(m)+a_(2)^(m)+...+a_(n)^(m))/(n)gt((a_(1)+a_(2)+...+a_(b))/(n))^(m)`

Options

A. 1
B. 2
C. 3
D. 4

Explanation

To solve the given problem, we will analyze both statements step by step.

### Statement 1:
We need to prove or disprove the statement:
\[ 1^3 + 3^3 + 5^3 + 7^3 + \ldots + (2n-1)^3 < n^4 \]

#### Step 1: Understanding the left-hand side
The left-hand side is the sum of cubes of the first \( n \) odd numbers. The formula for the sum of the cubes of the first \( n \) odd numbers is:
\[ \left( \frac{n(2n-1)}{2} \right)^2 \]

#### Step 2: Simplifying the expression
Using the formula for the sum of the first \( n \) odd numbers:
\[ 1 + 3 + 5 + \ldots + (2n-1) = n^2 \]
Thus, the sum of cubes becomes:
\[ (n^2)^2 = n^4 \]

#### Step 3: Comparing with \( n^4 \)
From the above, we see that:
\[ 1^3 + 3^3 + 5^3 + \ldots + (2n-1)^3 = n^4 \]
This means that:
\[ n^4 < n^4 \]
is not true. Therefore, Statement 1 is false.

### Statement 2:
We need to analyze the statement:
If \( a_1, a_2, \ldots, a_n \) are distinct positive real numbers and \( m > 1 \), then:
\[ \frac{a_1^m + a_2^m + \ldots + a_n^m}{n} > \left( \frac{a_1 + a_2 + \ldots + a_n}{n} \right)^m \]

#### Step 4: Applying Jensen's Inequality
Since \( f(x) = x^m \) is a convex function for \( m > 1 \), we can apply Jensen's inequality:
\[ f\left( \frac{a_1 + a_2 + \ldots + a_n}{n} \right) < \frac{f(a_1) + f(a_2) + \ldots + f(a_n)}{n} \]

This leads to:
\[ \left( \frac{a_1 + a_2 + \ldots + a_n}{n} \right)^m < \frac{a_1^m + a_2^m + \ldots + a_n^m}{n} \]

#### Step 5: Conclusion for Statement 2
Thus, Statement 2 is true.

### Final Conclusion:
- Statement 1 is false.
- Statement 2 is true.