The value of the sum `Sigma_(i=1)^(20) i(1/i+1/(i+1)+1/(i+2)+.....+1/(2))` is ______________.

Explanation

To solve the problem, we need to evaluate the sum:

\[ S = \sum_{i=1}^{20} i \left( \frac{1}{i} + \frac{1}{i+1} + \frac{1}{i+2} + \ldots + \frac{1}{20} \right) \]

### Step 1: Rewrite the Inner Sum
The inner sum can be rewritten as:

\[ \sum_{j=i}^{20} \frac{1}{j} \]

Thus, we can express \( S \) as:

\[ S = \sum_{i=1}^{20} i \sum_{j=i}^{20} \frac{1}{j} \]

### Step 2: Change the Order of Summation
We can change the order of summation. The term \( i \) will contribute to all \( j \) from \( i \) to \( 20 \). Therefore, we can express \( S \) as:

\[ S = \sum_{j=1}^{20} \frac{1}{j} \sum_{i=1}^{j} i \]

### Step 3: Evaluate the Inner Sum
The inner sum \( \sum_{i=1}^{j} i \) is the sum of the first \( j \) natural numbers, which can be calculated using the formula:

\[ \sum_{i=1}^{j} i = \frac{j(j+1)}{2} \]

### Step 4: Substitute Back into the Sum
Substituting this back into our expression for \( S \):

\[ S = \sum_{j=1}^{20} \frac{1}{j} \cdot \frac{j(j+1)}{2} \]

This simplifies to:

\[ S = \sum_{j=1}^{20} \frac{j+1}{2} \]

### Step 5: Simplify the Sum
We can separate the sum:

\[ S = \frac{1}{2} \sum_{j=1}^{20} (j + 1) = \frac{1}{2} \left( \sum_{j=1}^{20} j + \sum_{j=1}^{20} 1 \right) \]

The first part \( \sum_{j=1}^{20} j \) is again the sum of the first 20 natural numbers:

\[ \sum_{j=1}^{20} j = \frac{20 \cdot 21}{2} = 210 \]

The second part \( \sum_{j=1}^{20} 1 \) simply counts the number of terms, which is 20.

### Step 6: Combine the Results
Now we can combine these results:

\[ S = \frac{1}{2} \left( 210 + 20 \right) = \frac{1}{2} \cdot 230 = 115 \]

### Final Answer
Thus, the value of the sum is:

\[ \boxed{115} \]