Sum of the series `overset(2016)underset(r=1)Sigma (-1)^(r )(a+rd)` is

Options

A. a + 2016d
B. 2015d
C. 0
D. 1008d

Answer

To solve the series \( \sum_{r=1}^{2016} (-1)^{r}(a + rd) \), we will break down the steps systematically.

### Step 1: Write out the series
The series can be expressed as:
\[ \sum_{r=1}^{2016} (-1)^{r}(a + rd) = (-1)^1(a + 1d) + (-1)^2(a + 2d) + (-1)^3(a + 3d) + \ldots + (-1)^{2016}(a + 2016d) \]
This expands to:
\[ -a - d + a + 2d - a - 3d + a + 4d - \ldots + a + 2016d \]

### Step 2: Group the terms
We can group the terms based on the parity of \( r \):
- For odd \( r \): \( - (a + d) - (a + 3d) - (a + 5d) - \ldots - (a + 2015d) \)
- For even \( r \): \( (a + 2d) + (a + 4d) + (a + 6d) + \ldots + (a + 2016d) \)

### Step 3: Count the number of terms
There are 2016 terms in total, which means:
- Odd terms: \( 1008 \)
- Even terms: \( 1008 \)

### Step 4: Write the sums of odd and even terms
The odd terms can be expressed as:
\[ -(1008a + (1 + 3 + 5 + \ldots + 2015)d) \]
The even terms can be expressed as:
\[ 1008a + (2 + 4 + 6 + \ldots + 2016)d \]

### Step 5: Calculate the sums of the series
1. **Sum of odd integers** from 1 to 2015:
The sum of the first \( n \) odd numbers is \( n^2 \). Here, \( n = 1008 \):
\[ 1 + 3 + 5 + \ldots + 2015 = 1008^2 \]

2. **Sum of even integers** from 2 to 2016:
The sum of the first \( n \) even numbers is \( n(n + 1) \). Here, \( n = 1008 \):
\[ 2 + 4 + 6 + \ldots + 2016 = 1008 \times 1009 \]

### Step 6: Substitute back into the equation
Now substituting these sums back into our expression:
\[ -(1008a + 1008^2 d) + (1008a + 1008 \times 1009 d) \]

### Step 7: Simplify the expression
The \( 1008a \) terms cancel out:
\[ -(1008^2 d) + (1008 \times 1009 d) = (1008 \times 1009 - 1008^2)d \]
This simplifies to:
\[ 1008 \times (1009 - 1008)d = 1008d \]

### Final Answer
Thus, the sum of the series is:
\[ \boxed{1008d} \]