The sum of the series `sum_(r=1)^(n) (-1)^(r-1).""^(n)C_(r)(a-r), n gt 1` is equal to :

Options

A. `n.2^(n+1)+a`
B. 0
C. a
D. None of these

Answer

To solve the series

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

we can break it down step by step.

### Step 1: Rewrite the series
We can express the series as:

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

This separates the series into two parts: one involving \(a\) and the other involving \(r\).

### Step 2: Evaluate the first part
The first part of the series is:

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

Using the binomial theorem, we know:

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

Setting \(x = -1\):

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

Thus,

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

So, the first part simplifies to:

\[ a \cdot 1 = a \]

### Step 3: Evaluate the second part
Now we need to evaluate:

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

We can rewrite \(r\) as \(r = n \cdot \frac{1}{n} \cdot \binom{n-1}{r-1}\):

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

Changing the index of summation (let \(k = r - 1\)) gives:

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

Using the binomial theorem again:

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

Thus, the second part simplifies to:

\[ n \cdot 0 = 0 \]

### Step 4: Combine the results
Putting it all together, we have:

\[ S = a - 0 = a \]

Thus, the sum of the series is:

\[ \boxed{a} \]