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 start with the given equation:

\[ \frac{1}{a_1 + \omega} + \frac{1}{a_2 + \omega} + \frac{1}{a_3 + \omega} + \ldots + \frac{1}{a_n + \omega} = i \]

where \( \omega \) is an imaginary cube root of unity. The cube roots of unity are \( 1, \omega, \omega^2 \) where \( \omega = -\frac{1}{2} + i\frac{\sqrt{3}}{2} \) and \( \omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \).

### Step 1: Rewrite the fractions

We can rewrite each term in the sum:

\[ \frac{1}{a_r + \omega} = \frac{\omega^2}{(a_r + \omega)(\omega^2)} = \frac{\omega^2}{a_r \omega^2 + 1} \]

This gives us:

\[ \frac{1}{a_r + \omega} = \frac{\omega^2}{a_r + \omega} \cdot \frac{a_r + \omega^2}{a_r + \omega^2} = \frac{a_r + \omega^2}{a_r^2 + a_r \omega + 1} \]

### Step 2: Substitute into the sum

Now substituting this back into the original equation, we have:

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

### Step 3: Separate real and imaginary parts

Since \( \omega^2 = -\frac{1}{2} - i\frac{\sqrt{3}}{2} \), we can separate the real and imaginary parts of the equation. The real part of \( i \) is \( 0 \) and the imaginary part is \( 1 \).

### Step 4: Analyze the sum

The sum can be split into real and imaginary parts:

\[ \sum_{r=1}^{n} \frac{a_r - \frac{1}{2}}{a_r^2 - a_r + 1} + i \sum_{r=1}^{n} \frac{-\frac{\sqrt{3}}{2}}{a_r^2 - a_r + 1} = i \]

From this, we can equate the real part to \( 0 \):

\[ \sum_{r=1}^{n} \frac{2a_r - 1}{a_r^2 - a_r + 1} = 0 \]

### Step 5: Conclusion

Thus, the value of the sum is:

\[ \sum_{r=1}^{n} \frac{2a_r - 1}{a_r^2 - a_r + 1} = 0 \]