If `omega` is a complex nth root of unity, then `sum_(r=1)^n(a+b)omega^(r-1)` is equal to `(n(n+1)a)/2` b. `(n b)/(1+n)` c. `(n a)/(omega-1)` d. none of these

Options

A. `(n(n+1)a)/(2omega)`
B. `(nb)/(1-n)`
C. `(na)/(omega-1)`
D. None of these

Answer

To solve the problem, we need to evaluate the sum:

\[ S = \sum_{r=1}^{n} (a + b) \omega^{r-1} \]

where \(\omega\) is a complex nth root of unity.

### Step-by-Step Solution:

1. **Factor Out the Constant**:
We can factor out the constant \((a + b)\) from the summation:

\[ S = (a + b) \sum_{r=1}^{n} \omega^{r-1} \]

2. **Recognize the Summation**:
The summation \(\sum_{r=1}^{n} \omega^{r-1}\) is a geometric series. The first term \(a_1 = 1\) (when \(r=1\), \(\omega^{0} = 1\)) and the common ratio \(r = \omega\).

3. **Use the Formula for the Sum of a Geometric Series**:
The formula for the sum of the first \(n\) terms of a geometric series is given by:

\[ S_n = \frac{a_1(1 - r^n)}{1 - r} \]

Applying this to our series:

\[ \sum_{r=1}^{n} \omega^{r-1} = \frac{1(1 - \omega^n)}{1 - \omega} \]

4. **Substituting the Value of \(\omega^n\)**:
Since \(\omega\) is an nth root of unity, we have \(\omega^n = 1\). Thus, the sum simplifies to:

\[ \sum_{r=1}^{n} \omega^{r-1} = \frac{1 - 1}{1 - \omega} = 0 \]

5. **Final Result**:
Therefore, substituting back into our expression for \(S\):

\[ S = (a + b) \cdot 0 = 0 \]

### Conclusion:
The sum \(\sum_{r=1}^{n} (a + b) \omega^{r-1}\) equals \(0\).

### Answer:
None of the options provided match this result, so the answer is **d. none of these**.