If the sum of first n natural numbers is one-fifth of the sum of their squares, then n is

Options

A. 5
B. 6
C. 7
D. 8

Answer

To solve the problem, we need to find the value of \( n \) such that the sum of the first \( n \) natural numbers is one-fifth of the sum of their squares.

### Step-by-Step Solution:

1. **Write the formula for the sum of the first \( n \) natural numbers:**
\[ S = \frac{n(n + 1)}{2} \]

2. **Write the formula for the sum of the squares of the first \( n \) natural numbers:**
\[ S' = \frac{n(n + 1)(2n + 1)}{6} \]

3. **Set up the equation based on the problem statement:**
According to the problem, the sum of the first \( n \) natural numbers is one-fifth of the sum of their squares:
\[ S = \frac{1}{5} S' \]
Substituting the formulas for \( S \) and \( S' \):
\[ \frac{n(n + 1)}{2} = \frac{1}{5} \cdot \frac{n(n + 1)(2n + 1)}{6} \]

4. **Multiply both sides by 30 to eliminate the fractions:**
\[ 30 \cdot \frac{n(n + 1)}{2} = 30 \cdot \frac{1}{5} \cdot \frac{n(n + 1)(2n + 1)}{6} \]
This simplifies to:
\[ 15n(n + 1) = n(n + 1)(2n + 1) \]

5. **Assuming \( n(n + 1) \neq 0 \) (which is true for \( n \geq 1 \)), divide both sides by \( n(n + 1) \):**
\[ 15 = 2n + 1 \]

6. **Solve for \( n \):**
Rearranging gives:
\[ 2n = 15 - 1 \]
\[ 2n = 14 \]
\[ n = 7 \]

### Conclusion:
Thus, the value of \( n \) is \( 7 \).