`sum_(r=1)^(n) (-1)^(r-1) ""^nC_r(a - r) = `

Options

A. a
B. `-a`
C. 2a
D. 3a

Answer

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

\[ \sum_{r=1}^{n} (-1)^{r-1} \binom{n}{r} (a - r) \]

We can break this summation into two parts:

1. **Part 1:** \(\sum_{r=1}^{n} (-1)^{r-1} \binom{n}{r} a\)
2. **Part 2:** \(\sum_{r=1}^{n} (-1)^{r-1} \binom{n}{r} r\)

Thus, we can express the original sum as:

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

### Step 1: Evaluate the first part

The first part can be simplified using the binomial theorem. Specifically, we know that:

\[ \sum_{r=0}^{n} (-1)^{r} \binom{n}{r} = (1 - 1)^n = 0 \]

Thus, we have:

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

So,

\[ S_1 = a \cdot 1 = a \]

### Step 2: Evaluate the second part

For the second part, we can use the identity:

\[ \sum_{r=1}^{n} (-1)^{r-1} \binom{n}{r} r = n \cdot (-1)^{n-1} \]

This can be derived from differentiating the binomial expansion. Therefore, we have:

\[ S_2 = n \cdot (-1)^{n-1} \]

### Step 3: Combine the results

Now, we can combine both parts to find the final result:

\[ S = a - n \cdot (-1)^{n-1} \]

Thus, the final answer is:

\[ \sum_{r=1}^{n} (-1)^{r-1} \binom{n}{r} (a - r) = a + n \cdot (-1)^{n-1} \]