If a `= Sigma_(n=0)^(oo) x^(n), b = Sigma_(n=0)^(oo) y^(n), c = Sigma__(n=0)^(oo) (xy)^(n) " Where " |x|, |y| lt 1`, then -

Options

A. `abc = a + b + c`
B. `ab + bc = ac + b`
C. `ac + bc + b`
D. `ab + ac = bc + a`

Answer

To solve the problem, we need to evaluate the sums \( a \), \( b \), and \( c \) given by:

\[ a = \sum_{n=0}^{\infty} x^n, \quad b = \sum_{n=0}^{\infty} y^n, \quad c = \sum_{n=0}^{\infty} (xy)^n \]

where \( |x| < 1 \) and \( |y| < 1 \).

### Step 1: Evaluate \( a \)

The series \( a \) is a geometric series with first term \( 1 \) (when \( n=0 \)) and common ratio \( x \). The sum of an infinite geometric series is given by the formula:

\[ S = \frac{a}{1 - r} \]

where \( a \) is the first term and \( r \) is the common ratio. Therefore, we have:

\[ a = \frac{1}{1 - x} \]

### Step 2: Evaluate \( b \)

Similarly, for \( b \):

\[ b = \sum_{n=0}^{\infty} y^n \]

This is also a geometric series with first term \( 1 \) and common ratio \( y \):

\[ b = \frac{1}{1 - y} \]

### Step 3: Evaluate \( c \)

Now, we evaluate \( c \):

\[ c = \sum_{n=0}^{\infty} (xy)^n \]

This is again a geometric series with first term \( 1 \) and common ratio \( xy \):

\[ c = \frac{1}{1 - xy} \]

### Step 4: Relate \( a \), \( b \), and \( c \)

From the results we have:

1. \( a = \frac{1}{1 - x} \)
2. \( b = \frac{1}{1 - y} \)
3. \( c = \frac{1}{1 - xy} \)

We can express \( x \) and \( y \) in terms of \( a \) and \( b \):

\[ 1 - x = \frac{1}{a} \implies x = 1 - \frac{1}{a} \]
\[ 1 - y = \frac{1}{b} \implies y = 1 - \frac{1}{b} \]

### Step 5: Substitute \( x \) and \( y \) into the expression for \( c \)

Now substituting \( x \) and \( y \) into the expression for \( c \):

\[ 1 - xy = 1 - \left(1 - \frac{1}{a}\right)\left(1 - \frac{1}{b}\right) \]

Expanding this:

\[ 1 - xy = 1 - \left(1 - \frac{1}{a} - \frac{1}{b} + \frac{1}{ab}\right) \]
\[ = \frac{1}{a} + \frac{1}{b} - \frac{1}{ab} \]

### Step 6: Set up the equation

From the expression for \( c \):

\[ c = \frac{1}{1 - xy} \]

Thus,

\[ 1 - xy = \frac{1}{c} \]

Equating both expressions for \( 1 - xy \):

\[ \frac{1}{c} = \frac{1}{a} + \frac{1}{b} - \frac{1}{ab} \]

### Step 7: Rearranging the equation

Multiplying through by \( abc \):

\[ ab = b + a - c \]

Rearranging gives us:

\[ bc + ac = ab + c \]

### Conclusion

Thus, we have derived the relationship:

\[ bc + ac = ab + c \]

This means the correct option is:

**Option C is true.**