If the sum of n natural numbers is one-third the sum of their cubes then the value of n is equal to

Options

A. 3
B. 5
C. 2
D. 4

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-third of the sum of their cubes.

### Step 1: Write the formula for the sum of the first \( n \) natural numbers.
The sum of the first \( n \) natural numbers is given by:
\[ S_n = \frac{n(n + 1)}{2} \]

### Step 2: Write the formula for the sum of the cubes of the first \( n \) natural numbers.
The sum of the cubes of the first \( n \) natural numbers is given by:
\[ S_{n, \text{cubes}} = \left( \frac{n(n + 1)}{2} \right)^2 \]

### Step 3: Set up the equation based on the problem statement.
According to the problem, the sum of the first \( n \) natural numbers is one-third of the sum of their cubes:
\[ S_n = \frac{1}{3} S_{n, \text{cubes}} \]
Substituting the formulas we have:
\[ \frac{n(n + 1)}{2} = \frac{1}{3} \left( \frac{n(n + 1)}{2} \right)^2 \]

### Step 4: Simplify the equation.
Multiply both sides by 6 to eliminate the fractions:
\[ 3n(n + 1) = \left( \frac{n(n + 1)}{2} \right)^2 \]
This simplifies to:
\[ 3n(n + 1) = \frac{n^2(n + 1)^2}{4} \]

### Step 5: Cross-multiply to eliminate the fraction.
Cross-multiplying gives:
\[ 12n(n + 1) = n^2(n + 1)^2 \]

### Step 6: Factor out common terms.
Assuming \( n(n + 1) \neq 0 \), we can divide both sides by \( n(n + 1) \):
\[ 12 = n(n + 1) \]

### Step 7: Rearrange the equation.
This leads to the quadratic equation:
\[ n^2 + n - 12 = 0 \]

### Step 8: Solve the quadratic equation.
We can factor this quadratic:
\[ (n - 3)(n + 4) = 0 \]
Thus, the solutions are:
\[ n = 3 \quad \text{or} \quad n = -4 \]

### Step 9: Determine the valid solution.
Since \( n \) must be a natural number, we discard \( n = -4 \). Therefore, we have:
\[ n = 3 \]

### Final Answer:
The value of \( n \) is \( 3 \).

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