Let `(1 + x + x^(2))^(n) = sum_(r=0)^(2n) a_(r) x^(r)` . If `sum_(r=0)^(2n)(1)/(a_(r))= alpha` , then ` sum_(r=0)^(2n) (r)/(a_(r))` =

Options

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

Explanation

To solve the problem step by step, let's analyze the given expression and derive the required result.

### Given:
\[ (1 + x + x^2)^n = \sum_{r=0}^{2n} a_r x^r \]
We need to find:
\[ \sum_{r=0}^{2n} \frac{r}{a_r} \]
given that:
\[ \sum_{r=0}^{2n} \frac{1}{a_r} = \alpha \]

### Step 1: Substitute \( x = \frac{1}{x} \)
We start by substituting \( x = \frac{1}{x} \) in the original expression:
\[ (1 + \frac{1}{x} + \frac{1}{x^2})^n = \left(\frac{x^2 + x + 1}{x^2}\right)^n = \frac{(x^2 + x + 1)^n}{x^{2n}} \]
This implies:
\[ (1 + x + x^2)^n = x^{2n} \cdot \sum_{r=0}^{2n} a_r x^r \]

### Step 2: Identify Coefficients
From the above equation, we can see that the coefficients of \( x^r \) in \( (1 + x + x^2)^n \) correspond to \( a_r \) and \( a_{2n - r} \). Thus:
\[ a_r = a_{2n - r} \]

### Step 3: Relate \( \sum_{r=0}^{2n} \frac{r}{a_r} \) and \( \sum_{r=0}^{2n} \frac{2n - r}{a_r} \)
Using the symmetry of the coefficients:
\[ \sum_{r=0}^{2n} \frac{r}{a_r} = \sum_{r=0}^{2n} \frac{2n - r}{a_{2n - r}} = \sum_{r=0}^{2n} \frac{2n}{a_r} - \sum_{r=0}^{2n} \frac{r}{a_r} \]
Let \( S = \sum_{r=0}^{2n} \frac{r}{a_r} \). Then we have:
\[ S = \sum_{r=0}^{2n} \frac{2n}{a_r} - S \]
This simplifies to:
\[ 2S = 2n \sum_{r=0}^{2n} \frac{1}{a_r} \]
Thus:
\[ S = n \sum_{r=0}^{2n} \frac{1}{a_r} \]

### Step 4: Substitute the value of \( \sum_{r=0}^{2n} \frac{1}{a_r} \)
From the given information:
\[ \sum_{r=0}^{2n} \frac{1}{a_r} = \alpha \]
Substituting this into the equation for \( S \):
\[ S = n \alpha \]

### Final Result
Thus, we conclude that:
\[ \sum_{r=0}^{2n} \frac{r}{a_r} = n \alpha \]