For any positive real number a and for any `n in N`, the greatest value of `(a^(n))/(1+a+a^(2)+…+a^(2n))`, is

Options

A. `(1)/(2n)`
B. `(1)/(2n+1)`
C. `(1)/(2n-1)`
D. none of these

Explanation

To find the greatest value of the expression \( \frac{a^n}{1 + a + a^2 + \ldots + a^{2n}} \) for any positive real number \( a \) and for any \( n \in \mathbb{N} \), we can follow these steps:

### Step 1: Rewrite the Denominator
The denominator \( 1 + a + a^2 + \ldots + a^{2n} \) is a geometric series. The sum of a geometric series can be calculated using the formula:
\[ S = \frac{a^m - 1}{a - 1} \text{ for } a \neq 1 \]
where \( m \) is the number of terms. Here, we have \( 2n + 1 \) terms, so:
\[ 1 + a + a^2 + \ldots + a^{2n} = \frac{a^{2n+1} - 1}{a - 1} \text{ for } a \neq 1 \]

### Step 2: Substitute into the Expression
Now substitute this back into the expression:
\[ \frac{a^n}{1 + a + a^2 + \ldots + a^{2n}} = \frac{a^n}{\frac{a^{2n+1} - 1}{a - 1}} = \frac{a^n (a - 1)}{a^{2n+1} - 1} \]

### Step 3: Analyze the Expression
To find the maximum value, we can apply the Arithmetic Mean-Geometric Mean (AM-GM) inequality. The AM-GM inequality states that the arithmetic mean of non-negative numbers is greater than or equal to their geometric mean.

### Step 4: Apply AM-GM Inequality
We can express \( 1 + a + a^2 + \ldots + a^{2n} \) in terms of AM-GM:
\[ \frac{1 + a + a^2 + \ldots + a^{2n}}{2n + 1} \geq \sqrt[2n + 1]{1 \cdot a \cdot a^2 \cdots a^{2n}} = \sqrt[2n + 1]{a^{1 + 2 + \ldots + 2n}} = \sqrt[2n + 1]{a^{n(2n + 1)}} \]
Thus, we have:
\[ \frac{1 + a + a^2 + \ldots + a^{2n}}{2n + 1} \geq a^n \]

### Step 5: Rearranging the Inequality
Rearranging gives:
\[ 1 + a + a^2 + \ldots + a^{2n} \geq (2n + 1) a^n \]
Thus:
\[ \frac{a^n}{1 + a + a^2 + \ldots + a^{2n}} \leq \frac{a^n}{(2n + 1)a^n} = \frac{1}{2n + 1} \]

### Step 6: Conclusion
The greatest value of the expression \( \frac{a^n}{1 + a + a^2 + \ldots + a^{2n}} \) is therefore:
\[ \frac{1}{2n + 1} \]

### Final Answer
The greatest value of \( \frac{a^n}{1 + a + a^2 + \ldots + a^{2n}} \) is \( \frac{1}{2n + 1} \).