Lines `x=n, y=n^(3)` intersect at the point `A_(n).L_(n)` is a line through `A_(n)` parallel to `x+y=0` and make an intercept of length `l_(n)` on the axis of x, then the value of `sum_(n=1)^(5)l_(n)` is equal to `K^(4)-16`, where K =

Answer

To solve the problem step by step, let's break down the process:

### Step 1: Identify the Intersection Point
The lines given are \(x = n\) and \(y = n^3\). The intersection point \(A_n\) can be found by substituting \(n\) into both equations.

- **Intersection Point**:
\[ A_n = (n, n^3) \]

### Step 2: Find the Slope of the Line Parallel to \(x + y = 0\)
The line \(x + y = 0\) can be rewritten as \(y = -x\), which has a slope of \(-1\). A line parallel to this will also have a slope of \(-1\).

### Step 3: Write the Equation of the Line Through \(A_n\)
Using the point-slope form of the line equation, where the point is \(A_n(n, n^3)\) and the slope is \(-1\):

- **Equation of the Line**:
\[ y - n^3 = -1(x - n) \]
Simplifying this gives:
\[ y = -x + n + n^3 \]

### Step 4: Find the X-Intercept of the Line
To find the x-intercept, set \(y = 0\):

\[ 0 = -x + n + n^3 \]
Solving for \(x\):
\[ x = n + n^3 \]

### Step 5: Calculate the Length of the Intercept \(l_n\)
The length of the intercept \(l_n\) on the x-axis is the absolute value of the x-intercept:

- **Length of the Intercept**:
\[ l_n = n + n^3 \]

### Step 6: Sum \(l_n\) from \(n = 1\) to \(5\)
We need to compute:
\[ \sum_{n=1}^{5} l_n = \sum_{n=1}^{5} (n + n^3) \]
This can be split into two separate sums:
\[ \sum_{n=1}^{5} n + \sum_{n=1}^{5} n^3 \]

### Step 7: Calculate Each Sum
1. **Sum of the first 5 natural numbers**:
\[ \sum_{n=1}^{5} n = \frac{5(5+1)}{2} = \frac{5 \times 6}{2} = 15 \]

2. **Sum of the cubes of the first 5 natural numbers**:
\[ \sum_{n=1}^{5} n^3 = \left(\frac{5(5+1)}{2}\right)^2 = 15^2 = 225 \]

### Step 8: Combine the Results
Now, combine the results of the two sums:
\[ \sum_{n=1}^{5} l_n = 15 + 225 = 240 \]

### Step 9: Relate to \(K\)
According to the problem, we have:
\[ \sum_{n=1}^{5} l_n = K^4 - 16 \]
Setting this equal to our calculated sum:
\[ 240 = K^4 - 16 \]
Solving for \(K^4\):
\[ K^4 = 240 + 16 = 256 \]
Taking the fourth root:
\[ K = 4 \]

### Final Answer
Thus, the value of \(K\) is:
\[ \boxed{4} \]