If sum of the mth powers of first n odd numbers is `lambda,Aamgt1`, then (A)`lambdaltn^(m)` (B)`lambdagtn^(m)` (C)`lambdaltn^(m+1)` (D)`lambdagtn^(m+1)`

Options

A. `lambdaltn^(m)`
B. `lambdagtn^(m)`
C. `lambdaltn^(m+1)`
D. `lambdagtn^(m+1)`

Explanation

To solve the problem, we need to find the relationship between the sum of the mth powers of the first n odd numbers, denoted as \( \lambda \), and the variables \( n \) and \( m \).

### Step-by-Step Solution:

1. **Understanding the Sum of Odd Numbers**:
The first n odd numbers are \( 1, 3, 5, \ldots, (2n - 1) \). The sum of the mth powers of these numbers can be expressed as:
\[ S = 1^m + 3^m + 5^m + \ldots + (2n - 1)^m \]
This sum is given to be equal to \( \lambda \).

2. **Using the Arithmetic Mean-Geometric Mean Inequality (AM-GM)**:
The AM-GM inequality states that for any set of positive numbers, the arithmetic mean is greater than or equal to the geometric mean. Thus, we can write:
\[ \frac{S}{n} \geq \sqrt[n]{1^m \cdot 3^m \cdot 5^m \cdots (2n - 1)^m} \]
This simplifies to:
\[ \frac{\lambda}{n} \geq \sqrt[n]{(1 \cdot 3 \cdot 5 \cdots (2n - 1))^m} \]

3. **Calculating the Product of Odd Numbers**:
The product of the first n odd numbers can be expressed using the double factorial:
\[ 1 \cdot 3 \cdot 5 \cdots (2n - 1) = \frac{(2n)!}{2^n n!} \]
Therefore, we can rewrite the geometric mean as:
\[ \sqrt[n]{(1 \cdot 3 \cdot 5 \cdots (2n - 1))^m} = \sqrt[n]{\left(\frac{(2n)!}{2^n n!}\right)^m} \]

4. **Simplifying the AM-GM Inequality**:
From the AM-GM inequality, we have:
\[ \frac{\lambda}{n} \geq \left(\frac{(2n)!}{2^n n!}\right)^{\frac{m}{n}} \]
This implies:
\[ \lambda \geq n \cdot \left(\frac{(2n)!}{2^n n!}\right)^{\frac{m}{n}} \]

5. **Finding the Relationship**:
As \( n \) becomes large, the factorial terms can be approximated using Stirling’s approximation, leading us to conclude that:
\[ \lambda \sim n^{m+1} \]
Therefore, we can express this relationship as:
\[ \lambda > n^{m+1} \]

6. **Conclusion**:
Thus, the correct relationship is:
\[ \lambda > n^{m+1} \]
Hence, the answer is option (D) \( \lambda > n^{m+1} \).