If `omega` is a complex nth root of unity, then `underset(r=1)overset(n)(ar+b)omega^(r-1)` is equal to

Options

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

Explanation

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

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

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

### Step 1: Rewrite the Sum
We can break down the sum into two separate sums:

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

### Step 2: Evaluate the Second Sum
The second sum can be simplified as follows:

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

Since \(\omega^n = 1\) (because \(\omega\) is an nth root of unity), we have:

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

### Step 3: Evaluate the First Sum
Now we focus on the first sum:

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

Let \(S_1 = \sum_{r=1}^{n} r \omega^{r-1}\).

### Step 4: Use the Formula for the Sum of a Geometric Series
We can derive a formula for \(S_1\) by multiplying it by \(\omega\):

\[ \omega S_1 = \sum_{r=1}^{n} r \omega^r \]

Now we subtract the two equations:

\[ S_1 - \omega S_1 = \sum_{r=1}^{n} r \omega^{r-1} - \sum_{r=1}^{n} r \omega^r \]

This simplifies to:

\[ S_1(1 - \omega) = 1 + 2\omega + 3\omega^2 + \ldots + n\omega^{n-1} - n\omega^n \]

Since \(\omega^n = 1\), we can rewrite this as:

\[ S_1(1 - \omega) = 1 + 2\omega + 3\omega^2 + \ldots + n\omega^{n-1} - n \]

### Step 5: Solve for \(S_1\)
Now we can express \(S_1\):

\[ S_1(1 - \omega) = \sum_{r=1}^{n} r \omega^{r-1} - n \]

Thus,

\[ S_1 = \frac{1 + 2\omega + 3\omega^2 + \ldots + n\omega^{n-1} - n}{1 - \omega} \]

### Final Result
Since the second sum is zero, the final result for the original sum is:

\[ S = S_1 + 0 = S_1 = \frac{1 + 2\omega + 3\omega^2 + \ldots + n\omega^{n-1} - n}{1 - \omega} \]