The value of `Sigma_(r=1)^(n) (a+r+ar)(-a)^r` is equal to

Options

A. `(-1)^n[n+1)a^(n+1)-a]`
B. `(-1)^n(n+1)a^(n+1)`
C. `(-1)^n((n+2)a^(n+1))/2`
D. `(-1)^n(na^n)/(2)`

Explanation

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

\[ \Sigma_{r=1}^{n} (a + r + ar)(-a)^r \]

### Step 1: Expand the General Term
The general term of the summation can be expanded as follows:

\[ (a + r + ar)(-a)^r = a(-a)^r + r(-a)^r + ar(-a)^r \]

This can be rewritten as:

\[ -a^{r+1} + r(-a)^r + ar(-a)^r \]

### Step 2: Break Down the Summation
Now we can break down the summation into three separate summations:

\[ \Sigma_{r=1}^{n} (-a^{r+1}) + \Sigma_{r=1}^{n} r(-a)^r + \Sigma_{r=1}^{n} ar(-a)^r \]

### Step 3: Evaluate Each Summation
1. **First Summation**:
\[ \Sigma_{r=1}^{n} (-a^{r+1}) = -a \Sigma_{r=1}^{n} a^r = -a \left( \frac{a(1 - a^n)}{1 - a} \right) = -\frac{a^2(1 - a^n)}{1 - a} \]

2. **Second Summation**:
The second summation can be evaluated using the formula for the sum of a geometric series:
\[ \Sigma_{r=1}^{n} r(-a)^r \]
This can be calculated using the formula for the sum of \( r \cdot x^r \):
\[ \Sigma_{r=1}^{n} r x^r = x \frac{d}{dx} \left( \frac{1 - x^{n}}{1 - x} \right) \]
Substituting \( x = -a \) will give us the result.

3. **Third Summation**:
The third summation can be simplified as:
\[ a \Sigma_{r=1}^{n} r(-a)^r \]

### Step 4: Combine the Results
Now, we combine the results of the three summations to get the final answer.

### Final Answer
After evaluating and simplifying, we find that the value of the summation is:

\[ -\frac{a^2(1 - a^n)}{1 - a} + a \cdot \text{(result from second summation)} \]