If `x=sum_(n=0)^(oo) a^(n),y=sum_(n=0)^(oo)b^(n),z=sum_(n=0)^(oo)(ab)^(n)`, where `a,blt1`, then

Options

A. xyz=x+y+z
B. xz+yz=xy+z
C. xy+yz=xz+y
D. xy+xz=yz+x

Explanation

To solve the problem, we need to find the relationship between \( x \), \( y \), and \( z \) given the definitions of these sums.

### Step 1: Define the sums
We have:
- \( x = \sum_{n=0}^{\infty} a^n \)
- \( y = \sum_{n=0}^{\infty} b^n \)
- \( z = \sum_{n=0}^{\infty} (ab)^n \)

### Step 2: Use the formula for the sum of a geometric series
The sum of an infinite geometric series can be expressed as:
\[ \sum_{n=0}^{\infty} r^n = \frac{1}{1 - r} \quad \text{for } |r| < 1 \]

Using this formula, we can express \( x \), \( y \), and \( z \) as follows:
- For \( x \):
\[ x = \frac{1}{1 - a} \quad \text{(since } |a| < 1\text{)} \]
- For \( y \):
\[ y = \frac{1}{1 - b} \quad \text{(since } |b| < 1\text{)} \]
- For \( z \):
\[ z = \frac{1}{1 - ab} \quad \text{(since } |ab| < 1\text{)} \]

### Step 3: Rearranging the equations
From the expressions for \( x \), \( y \), and \( z \), we can rearrange them:
- From \( x = \frac{1}{1 - a} \):
\[ 1 - a = \frac{1}{x} \implies a = 1 - \frac{1}{x} \]
- From \( y = \frac{1}{1 - b} \):
\[ 1 - b = \frac{1}{y} \implies b = 1 - \frac{1}{y} \]
- From \( z = \frac{1}{1 - ab} \):
\[ 1 - ab = \frac{1}{z} \implies ab = 1 - \frac{1}{z} \]

### Step 4: Substitute \( a \) and \( b \) into the equation for \( ab \)
Now we substitute the expressions for \( a \) and \( b \) into the equation for \( ab \):
\[ ab = \left(1 - \frac{1}{x}\right)\left(1 - \frac{1}{y}\right) \]

### Step 5: Expand the product
Expanding the product gives:
\[ ab = 1 - \frac{1}{x} - \frac{1}{y} + \frac{1}{xy} \]

### Step 6: Set up the equation with \( z \)
From our earlier expression for \( ab \):
\[ 1 - ab = \frac{1}{z} \]
Substituting \( ab \) from the expansion:
\[ 1 - \left(1 - \frac{1}{x} - \frac{1}{y} + \frac{1}{xy}\right) = \frac{1}{z} \]
This simplifies to:
\[ \frac{1}{x} + \frac{1}{y} - \frac{1}{xy} = \frac{1}{z} \]

### Step 7: Rearranging to find the relationship
Multiplying through by \( xyz \) gives:
\[ yz + xz - xy = xyz \]
Rearranging this gives us the final relationship:
\[ xy + xz + yz = xyz \]

### Conclusion
Thus, the relationship between \( x \), \( y \), and \( z \) is:
\[ xy + xz + yz = xyz \]