Find `F_("net") = GMm[ (1)/(r^(2))+ (1)/(2r^(2))+ (1)/(4r^(2))+ ...... "up to "oo]`.

Explanation

To find the net force \( F_{\text{net}} \), we need to sum the given infinite series. The series given is:

\[ F_{\text{net}} = GMm \left( \frac{1}{r^2} + \frac{1}{2r^2} + \frac{1}{4r^2} + \cdots \right) \]

This is a geometric progression (GP) series. Let's solve it step by step.

### Step 1: Identify the first term (a) and the common ratio (r)
The given series is:

\[ \frac{1}{r^2} + \frac{1}{2r^2} + \frac{1}{4r^2} + \cdots \]

Here, the first term \( a \) is:

\[ a = \frac{1}{r^2} \]

To find the common ratio \( r \), divide the second term by the first term:

\[ r = \frac{\frac{1}{2r^2}}{\frac{1}{r^2}} = \frac{1}{2} \]

### Step 2: Use the sum formula for an infinite geometric series
The sum \( S \) of an infinite geometric series with first term \( a \) and common ratio \( r \) (where \( |r| < 1 \)) is given by:

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

### Step 3: Substitute the values of \( a \) and \( r \) into the sum formula
Here, \( a = \frac{1}{r^2} \) and \( r = \frac{1}{2} \):

\[ S = \frac{\frac{1}{r^2}}{1 - \frac{1}{2}} \]

### Step 4: Simplify the expression
Simplify the denominator:

\[ 1 - \frac{1}{2} = \frac{1}{2} \]

So the sum \( S \) becomes:

\[ S = \frac{\frac{1}{r^2}}{\frac{1}{2}} = \frac{1}{r^2} \cdot \frac{2}{1} = \frac{2}{r^2} \]

### Step 5: Multiply by the constant factor \( GMm \)
The net force \( F_{\text{net}} \) is:

\[ F_{\text{net}} = GMm \cdot S = GMm \cdot \frac{2}{r^2} = \frac{2GMm}{r^2} \]

### Final Answer
\[ F_{\text{net}} = \frac{2GMm}{r^2} \]