Find the value of `{:(" "SigmaSigma),(1 le i le j):} " "i xx (1/2)^j`

Explanation

To solve the problem of finding the value of the double summation \(\sum_{1 \leq i \leq j} i \left(\frac{1}{2}\right)^j\), we can follow these steps:

### Step 1: Rewrite the Double Summation
We start with the double summation:
\[ S = \sum_{j=1}^{\infty} \sum_{i=1}^{j} i \left(\frac{1}{2}\right)^j \]
This means we will first sum over \(i\) from 1 to \(j\), and then sum over \(j\).

### Step 2: Evaluate the Inner Summation
The inner summation \(\sum_{i=1}^{j} i\) can be calculated using the formula for the sum of the first \(n\) natural numbers:
\[ \sum_{i=1}^{j} i = \frac{j(j+1)}{2} \]
Thus, we can rewrite \(S\) as:
\[ S = \sum_{j=1}^{\infty} \frac{j(j+1)}{2} \left(\frac{1}{2}\right)^j \]

### Step 3: Factor Out the Constant
We can factor out \(\frac{1}{2}\):
\[ S = \frac{1}{2} \sum_{j=1}^{\infty} j(j+1) \left(\frac{1}{2}\right)^j \]

### Step 4: Simplify the Series
We can express \(j(j+1)\) as \(j^2 + j\):
\[ S = \frac{1}{2} \left( \sum_{j=1}^{\infty} j^2 \left(\frac{1}{2}\right)^j + \sum_{j=1}^{\infty} j \left(\frac{1}{2}\right)^j \right) \]

### Step 5: Evaluate Each Series
1. **For \(\sum_{j=1}^{\infty} j \left(\frac{1}{2}\right)^j\)**:
This is a standard result:
\[ \sum_{j=1}^{\infty} jx^j = \frac{x}{(1-x)^2} \quad \text{for } |x| < 1 \]
Setting \(x = \frac{1}{2}\):
\[ \sum_{j=1}^{\infty} j \left(\frac{1}{2}\right)^j = \frac{\frac{1}{2}}{\left(1 - \frac{1}{2}\right)^2} = \frac{\frac{1}{2}}{\left(\frac{1}{2}\right)^2} = 2 \]

2. **For \(\sum_{j=1}^{\infty} j^2 \left(\frac{1}{2}\right)^j\)**:
Using the formula:
\[ \sum_{j=1}^{\infty} j^2 x^j = \frac{x(1+x)}{(1-x)^3} \]
Setting \(x = \frac{1}{2}\):
\[ \sum_{j=1}^{\infty} j^2 \left(\frac{1}{2}\right)^j = \frac{\frac{1}{2}(1+\frac{1}{2})}{\left(1 - \frac{1}{2}\right)^3} = \frac{\frac{1}{2} \cdot \frac{3}{2}}{\left(\frac{1}{2}\right)^3} = \frac{\frac{3}{4}}{\frac{1}{8}} = 6 \]

### Step 6: Combine the Results
Now we can combine the results:
\[ S = \frac{1}{2} \left(6 + 2\right) = \frac{1}{2} \cdot 8 = 4 \]

### Final Answer
Thus, the value of the double summation is:
\[ \boxed{4} \]