For some natural number 'n', the sum of the first 'n' natural numbers is 240 less than the sum of the first `(n+5)` natural numbers. Then n itself is the sum of how many natural numbers starting with 1.

Answer

To solve the problem step by step, we will follow the logic presented in the video transcript.

### Step 1: Understand the Problem
We need to find a natural number \( n \) such that the sum of the first \( n \) natural numbers is 240 less than the sum of the first \( n + 5 \) natural numbers.

### Step 2: Write the Formula for the Sum of Natural Numbers
The sum of the first \( n \) natural numbers is given by the formula:
\[ S_n = \frac{n(n + 1)}{2} \]
The sum of the first \( n + 5 \) natural numbers is:
\[ S_{n+5} = \frac{(n + 5)(n + 6)}{2} \]

### Step 3: Set Up the Equation
According to the problem, we have:
\[ S_n + 240 = S_{n+5} \]
Substituting the formulas we have:
\[ \frac{n(n + 1)}{2} + 240 = \frac{(n + 5)(n + 6)}{2} \]

### Step 4: Eliminate the Denominator
To eliminate the fraction, multiply the entire equation by 2:
\[ n(n + 1) + 480 = (n + 5)(n + 6) \]

### Step 5: Expand Both Sides
Expanding both sides gives:
\[ n^2 + n + 480 = n^2 + 11n + 30 \]

### Step 6: Simplify the Equation
Now, we can simplify by subtracting \( n^2 \) from both sides:
\[ n + 480 = 11n + 30 \]
Rearranging gives:
\[ 480 - 30 = 11n - n \]
\[ 450 = 10n \]

### Step 7: Solve for \( n \)
Dividing both sides by 10:
\[ n = 45 \]

### Step 8: Find How Many Natural Numbers Sum to \( n \)
Now we need to find how many natural numbers sum to \( n \) (which is 45). The sum of the first \( m \) natural numbers is given by:
\[ \frac{m(m + 1)}{2} = 45 \]
Multiplying both sides by 2:
\[ m(m + 1) = 90 \]

### Step 9: Solve the Quadratic Equation
This can be rewritten as:
\[ m^2 + m - 90 = 0 \]
To solve for \( m \), we can factor the equation:
\[ (m - 9)(m + 10) = 0 \]
Thus, \( m = 9 \) (since \( m \) must be a natural number).

### Final Answer
Therefore, \( n \) itself is the sum of **9** natural numbers starting with 1.