If `(e^(x))/(1-x) = B_(0) +B_(1)x+B_(2)x^(2)+...+B_(n)x^(n)+... `, then the value of `B_(n) - B_(n-1)` is

Options

A. `(1)/(n!)`
B. `(1)/((n-1)!)`
C. `(1)/(n!) - (1)/((n -1)!)`
D. 1

Answer

To solve the problem, we start with the equation given:

\[ \frac{e^x}{1-x} = B_0 + B_1 x + B_2 x^2 + \ldots + B_n x^n + \ldots \]

### Step 1: Rewrite the equation

We can multiply both sides of the equation by \(1 - x\):

\[ e^x = (1 - x)(B_0 + B_1 x + B_2 x^2 + \ldots) \]

### Step 2: Expand the right-hand side

Expanding the right-hand side gives:

\[ e^x = B_0(1 - x) + B_1 x(1 - x) + B_2 x^2(1 - x) + \ldots \]

This can be simplified to:

\[ e^x = B_0 - B_0 x + B_1 x - B_1 x^2 + B_2 x^2 - B_2 x^3 + \ldots \]

### Step 3: Collect like terms

Now, we can collect the coefficients of \(x^n\):

- The constant term (coefficient of \(x^0\)) is \(B_0\).
- The coefficient of \(x\) is \(B_1 - B_0\).
- The coefficient of \(x^2\) is \(B_2 - B_1\).
- The coefficient of \(x^3\) is \(B_3 - B_2\).
- In general, the coefficient of \(x^n\) is \(B_n - B_{n-1}\).

### Step 4: Compare with the series expansion of \(e^x\)

The series expansion of \(e^x\) is:

\[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

### Step 5: Equate coefficients

Now we can equate the coefficients from both sides:

1. From the constant term:
\[ B_0 = 1 \]

2. From the coefficient of \(x\):
\[ B_1 - B_0 = 1 \implies B_1 - 1 = 1 \implies B_1 = 2 \]

3. From the coefficient of \(x^2\):
\[ B_2 - B_1 = \frac{1}{2!} \implies B_2 - 2 = \frac{1}{2} \implies B_2 = 2.5 \]

4. From the coefficient of \(x^3\):
\[ B_3 - B_2 = \frac{1}{3!} \implies B_3 - 2.5 = \frac{1}{6} \implies B_3 = 2.5 + \frac{1}{6} \]

### Step 6: Generalize the formula

Continuing this process, we find that:

\[ B_n - B_{n-1} = \frac{1}{n!} \]

### Conclusion

Thus, the value of \(B_n - B_{n-1}\) is:

\[ \boxed{\frac{1}{n!}} \]