If the sum of first `n` terms of an `A P` is `c n^2,` then the sum of squares of these `n` terms is (2009) `(n(4n^2-1)c^2)/6` (b) `(n(4n^2+1)c^2)/3` `(n(4n^2-1)c^2)/3` (d) `(n(4n^2+1)c^2)/6`

Options

A. `(n(4n^2-1)c^2)/6`
B. `(n(4n^2+1)c^2)/3`
C. `(n(4n^2-1)c^2)/3`
D. `(n(4n^2+1)c^2)/6`

Answer

To solve the problem, we need to find the sum of the squares of the first `n` terms of an arithmetic progression (AP) given that the sum of the first `n` terms is \( S_n = c n^2 \).

### Step-by-Step Solution:

1. **Identify the nth term of the AP**:
The nth term \( T_n \) can be found using the formula:
\[ T_n = S_n - S_{n-1} \]
Here, \( S_n = c n^2 \) and \( S_{n-1} = c(n-1)^2 \).

2. **Calculate \( S_{n-1} \)**:
\[ S_{n-1} = c(n-1)^2 = c(n^2 - 2n + 1) = c n^2 - 2cn + c \]

3. **Find \( T_n \)**:
Now, substituting \( S_n \) and \( S_{n-1} \) into the formula for \( T_n \):
\[ T_n = S_n - S_{n-1} = c n^2 - (c n^2 - 2cn + c) \]
Simplifying this gives:
\[ T_n = 2cn - c = c(2n - 1) \]

4. **Sum of squares of the first n terms**:
We need to find \( \sum_{k=1}^{n} T_k^2 \):
\[ \sum_{k=1}^{n} T_k^2 = \sum_{k=1}^{n} (c(2k - 1))^2 = c^2 \sum_{k=1}^{n} (2k - 1)^2 \]

5. **Expand \( (2k - 1)^2 \)**:
\[ (2k - 1)^2 = 4k^2 - 4k + 1 \]
Therefore,
\[ \sum_{k=1}^{n} (2k - 1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) = 4\sum_{k=1}^{n} k^2 - 4\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

6. **Use formulas for summation**:
We know:
- \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)
- \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)
- \( \sum_{k=1}^{n} 1 = n \)

Plugging these into our equation:
\[ \sum_{k=1}^{n} (2k - 1)^2 = 4 \cdot \frac{n(n+1)(2n+1)}{6} - 4 \cdot \frac{n(n+1)}{2} + n \]

7. **Simplify the expression**:
\[ = \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n \]
\[ = \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3} \]
\[ = \frac{2n(n+1)(2n+1 - 3) + 3n}{3} \]
\[ = \frac{2n(n+1)(2n-2) + 3n}{3} \]
\[ = \frac{2n(n+1)(2(n-1)) + 3n}{3} \]
\[ = \frac{4n(n^2 - 1)c^2 + 3n}{3} \]

8. **Final expression**:
Thus, the sum of the squares of the first `n` terms is:
\[ S' = c^2 \cdot \frac{n(4n^2 - 1)}{3} \]

### Conclusion:
The correct answer is:
\[ \frac{n(4n^2 - 1)c^2}{3} \]