If `sum_(r=0)^(n){a_(r)(x-alpha+2)^(r)-b_(r)(alpha-x-1)^(r)}=0`, then (a) `b_(n) = 1+a_(n)` (b) `b_(n) = (-1)^(n)xxa_(n)` (c) `b_(n) = (-1)^(n-1) xxa_(n)` (d) `b_(n) + 1 = a_(n)`

Options

A. `b_(n) = 1+a_(n)`
B. `b_(n) = (-1)^(n)xxa_(n)`
C. `b_(n) = (-1)^(n-1) xxa_(n)`
D. `b_(n) + 1 = a_(n)`

Explanation

To solve the problem, we need to analyze the given equation:

\[ \sum_{r=0}^{n} a_r (x - \alpha + 2)^r - b_r (\alpha - x - 1)^r = 0 \]

This implies that the two sums must be equal for all values of \(x\). Let's denote \(t = \alpha - x - 1\). Then we can rewrite \(x - \alpha + 2\) as follows:

\[ x - \alpha + 2 = 1 - t \]

Substituting this into the original equation gives:

\[ \sum_{r=0}^{n} a_r (1 - t)^r - b_r t^r = 0 \]

This can be rearranged to:

\[ \sum_{r=0}^{n} a_r (1 - t)^r = \sum_{r=0}^{n} b_r t^r \]

Now, we need to find the coefficient of \(t^n\) on both sides of the equation.

### Step 1: Expand the left-hand side

The left-hand side can be expressed as:

\[ \sum_{r=0}^{n} a_r (1 - t)^r = a_0 (1 - t)^0 + a_1 (1 - t)^1 + a_2 (1 - t)^2 + \ldots + a_n (1 - t)^n \]

### Step 2: Use the Binomial Theorem

Using the Binomial Theorem, we can expand \((1 - t)^r\):

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

Thus, we can rewrite the left-hand side as:

\[ \sum_{r=0}^{n} a_r \sum_{k=0}^{r} \binom{r}{k} (-t)^k \]

### Step 3: Collect terms for \(t^n\)

To find the coefficient of \(t^n\), we need to consider the contributions from each \(a_r\):

The coefficient of \(t^n\) in the left-hand side will be:

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

### Step 4: Right-hand side

For the right-hand side, we have:

\[ \sum_{r=0}^{n} b_r t^r \]

The coefficient of \(t^n\) here is simply \(b_n\).

### Step 5: Set the coefficients equal

Since both sides of the equation must be equal, we have:

\[ b_n = (-1)^n a_n \]

### Conclusion

Thus, the relationship between \(b_n\) and \(a_n\) is:

\[ b_n = (-1)^n a_n \]

Now, let's check the options given in the question:

- (a) \(b_n = 1 + a_n\)
- (b) \(b_n = (-1)^n a_n\) **(This is correct)**
- (c) \(b_n = (-1)^{n-1} a_n\)
- (d) \(b_n + 1 = a_n\)

The correct answer is option (b).