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 given equation:

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

### Step 1: Rearranging the equation
We can rearrange the equation to isolate \( e^x \):

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

### Step 2: Expanding \( e^x \)
The Taylor series expansion for \( e^x \) is:

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

### Step 3: Expanding the right-hand side
Now, we expand the right-hand side:

\[ (1 - x)(B_0 + B_1 x + B_2 x^2 + \ldots) = B_0 + (B_1 - B_0)x + (B_2 - B_1)x^2 + (B_3 - B_2)x^3 + \ldots \]

### Step 4: Comparing coefficients
Now we need to compare coefficients of \( x^n \) from both sides of the equation.

From the left-hand side, the coefficient of \( x^n \) in \( e^x \) is \( \frac{1}{n!} \).

From the right-hand side, the coefficient of \( x^n \) is \( B_n - B_{n-1} \).

### Step 5: Setting up the equation
Equating the coefficients gives us:

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

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

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

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