If `1+2+3+……+n=28`, then the value of `1^(2)+2^(2)+3^(2)+………..+n^(2)` is
(i) 560
(ii) 280
(iii) 140
(iv) None of these

Options

A. 560
B. 280
C. 140
D. None of these

Answer

To solve the problem, we need to find the value of \(1^2 + 2^2 + 3^2 + \ldots + n^2\) given that \(1 + 2 + 3 + \ldots + n = 28\).

### Step 1: Use the formula for the sum of the first \(n\) natural numbers.
The sum of the first \(n\) natural numbers is given by the formula:
\[ S_n = \frac{n(n + 1)}{2} \]
We know from the problem statement that:
\[ S_n = 28 \]
So we can set up the equation:
\[ \frac{n(n + 1)}{2} = 28 \]

### Step 2: Solve for \(n\).
Multiplying both sides by 2 to eliminate the fraction:
\[ n(n + 1) = 56 \]
Rearranging gives us:
\[ n^2 + n - 56 = 0 \]

### Step 3: Factor the quadratic equation.
We need to factor \(n^2 + n - 56\). We look for two numbers that multiply to \(-56\) and add to \(1\). The numbers \(8\) and \(-7\) work:
\[ (n - 7)(n + 8) = 0 \]
Setting each factor to zero gives:
\[ n - 7 = 0 \quad \text{or} \quad n + 8 = 0 \]
Thus, we have:
\[ n = 7 \quad \text{or} \quad n = -8 \]
Since \(n\) must be a natural number, we take:
\[ n = 7 \]

### Step 4: Use the formula for the sum of squares.
Now we need to find \(1^2 + 2^2 + 3^2 + \ldots + n^2\). The formula for the sum of the squares of the first \(n\) natural numbers is:
\[ S_{n^2} = \frac{n(2n + 1)(n + 1)}{6} \]
Substituting \(n = 7\):
\[ S_{n^2} = \frac{7(2 \cdot 7 + 1)(7 + 1)}{6} \]
Calculating inside the parentheses:
\[ = \frac{7(14 + 1)(8)}{6} = \frac{7 \cdot 15 \cdot 8}{6} \]

### Step 5: Simplify the expression.
Now we simplify:
\[ = \frac{7 \cdot 15 \cdot 8}{6} \]
Breaking it down:
\[ = \frac{7 \cdot 15 \cdot 4}{3} \quad \text{(since } 8/2 = 4 \text{ and } 6/3 = 2\text{)} \]
Now calculate:
\[ = \frac{7 \cdot 15 \cdot 4}{3} = \frac{420}{3} = 140 \]

### Final Answer:
Thus, the value of \(1^2 + 2^2 + 3^2 + \ldots + n^2\) is \(140\).