The difference between the sum of the first k terms of the series `1^3+2^3+3^3+....+n^3` and the sum of the first k terms of `1+2+3+.....+n` is `1980` . The value of k is :

Explanation

To solve the problem, we need to find the value of \( k \) such that the difference between the sum of the first \( k \) terms of the series \( 1^3 + 2^3 + 3^3 + \ldots + k^3 \) and the sum of the first \( k \) terms of the series \( 1 + 2 + 3 + \ldots + k \) is equal to \( 1980 \).

### Step 1: Write the formulas for the sums

The sum of the first \( k \) cubes is given by the formula:
\[ S_k = \left( \frac{k(k + 1)}{2} \right)^2 \]
The sum of the first \( k \) natural numbers is given by the formula:
\[ T_k = \frac{k(k + 1)}{2} \]

### Step 2: Set up the equation

According to the problem, the difference between these two sums is \( 1980 \):
\[ S_k - T_k = 1980 \]
Substituting the formulas for \( S_k \) and \( T_k \):
\[ \left( \frac{k(k + 1)}{2} \right)^2 - \frac{k(k + 1)}{2} = 1980 \]

### Step 3: Simplify the equation

Let \( x = \frac{k(k + 1)}{2} \). Then we can rewrite the equation as:
\[ x^2 - x = 1980 \]
This simplifies to:
\[ x^2 - x - 1980 = 0 \]

### Step 4: Solve the quadratic equation

Using the quadratic formula \( x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ x = \frac{1 \pm \sqrt{1 + 4 \cdot 1980}}{2} \]
Calculating the discriminant:
\[ 1 + 7920 = 7921 \]
Now, taking the square root:
\[ \sqrt{7921} = 89 \]
Thus, we have:
\[ x = \frac{1 \pm 89}{2} \]
Calculating the two possible values for \( x \):
1. \( x = \frac{90}{2} = 45 \)
2. \( x = \frac{-88}{2} = -44 \) (not valid since \( x \) must be positive)

So, \( x = 45 \).

### Step 5: Relate \( x \) back to \( k \)

Recall that \( x = \frac{k(k + 1)}{2} \):
\[ \frac{k(k + 1)}{2} = 45 \]
Multiplying both sides by 2:
\[ k(k + 1) = 90 \]

### Step 6: Solve for \( k \)

This gives us the quadratic equation:
\[ k^2 + k - 90 = 0 \]
Using the quadratic formula again:
\[ k = \frac{-1 \pm \sqrt{1 + 360}}{2} \]
Calculating the discriminant:
\[ 1 + 360 = 361 \]
Taking the square root:
\[ \sqrt{361} = 19 \]
Thus, we have:
\[ k = \frac{-1 \pm 19}{2} \]
Calculating the two possible values for \( k \):
1. \( k = \frac{18}{2} = 9 \)
2. \( k = \frac{-20}{2} = -10 \) (not valid since \( k \) must be positive)

### Final Answer

Thus, the value of \( k \) is:
\[ \boxed{9} \]