Let `1/(a_1+omega) + 1/(a_2+omega)+1/(a_3+omega)+ … . + 1/(a_n+omega)=i`
where `a_1,a_2,a_3` …. `a_n in R` and `omega` is imaginary cube root of unity , then evaluate `sum_(r=1)^(n)(2a_r-1)/(a_r^2-a_r+1)` .

Answer

To solve the problem, we need to evaluate the expression

\[ \sum_{r=1}^{n} \frac{2a_r - 1}{a_r^2 - a_r + 1} \]

given that

\[ \frac{1}{a_1 + \omega} + \frac{1}{a_2 + \omega} + \cdots + \frac{1}{a_n + \omega} = i \]

where \( \omega \) is the imaginary cube root of unity.

### Step-by-step Solution:

1. **Understanding the properties of \( \omega \)**:
The imaginary cube roots of unity are given by:
\[ \omega = e^{2\pi i / 3} = -\frac{1}{2} + \frac{\sqrt{3}}{2}i \]
and
\[ \omega^2 = e^{-2\pi i / 3} = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \]
We also know that:
\[ \omega + \omega^2 = -1 \quad \text{and} \quad \omega \cdot \omega^2 = 1 \]

2. **Rewriting the given sum**:
We can rewrite each term in the sum:
\[ \frac{1}{a_r + \omega} = \frac{1}{a_r + \omega} \cdot \frac{a_r + \omega^2}{a_r + \omega^2} = \frac{a_r + \omega^2}{(a_r + \omega)(a_r + \omega^2)} \]
The denominator simplifies to:
\[ (a_r + \omega)(a_r + \omega^2) = a_r^2 + a_r(\omega + \omega^2) + \omega \cdot \omega^2 = a_r^2 - a_r + 1 \]
Therefore, we have:
\[ \frac{1}{a_r + \omega} = \frac{a_r + \omega^2}{a_r^2 - a_r + 1} \]

3. **Summing the rewritten terms**:
Thus, the sum becomes:
\[ \sum_{r=1}^{n} \frac{a_r + \omega^2}{a_r^2 - a_r + 1} = i \]

4. **Separating real and imaginary parts**:
Substitute \( \omega^2 = -\frac{1}{2} - \frac{\sqrt{3}}{2}i \):
\[ \sum_{r=1}^{n} \frac{a_r - \frac{1}{2} - \frac{\sqrt{3}}{2}i}{a_r^2 - a_r + 1} = i \]
This can be separated into real and imaginary parts:
\[ \sum_{r=1}^{n} \frac{a_r - \frac{1}{2}}{a_r^2 - a_r + 1} + i \sum_{r=1}^{n} \frac{-\frac{\sqrt{3}}{2}}{a_r^2 - a_r + 1} = i \]

5. **Setting the real part to zero**:
The real part must equal zero:
\[ \sum_{r=1}^{n} \frac{a_r - \frac{1}{2}}{a_r^2 - a_r + 1} = 0 \]

6. **Relating to the desired sum**:
Now, we can express the desired sum:
\[ \sum_{r=1}^{n} \frac{2a_r - 1}{a_r^2 - a_r + 1} = 2 \sum_{r=1}^{n} \frac{a_r - \frac{1}{2}}{a_r^2 - a_r + 1} = 2 \cdot 0 = 0 \]

### Final Answer:
Thus, the value of the sum is:
\[ \sum_{r=1}^{n} \frac{2a_r - 1}{a_r^2 - a_r + 1} = 0 \]