If the sum of first n terms of an AP is `cn^(2)`, then the sum of squares of these n terms is

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)`

Explanation

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 \( S_n = cn^2 \).

### Step 1: Find the nth term of the AP
The nth term \( a_n \) of an AP can be expressed as:
\[ a_n = S_n - S_{n-1} \]
where \( S_n \) is the sum of the first n terms and \( S_{n-1} \) is the sum of the first \( n-1 \) terms.

Given:
\[ S_n = cn^2 \]
We can find \( S_{n-1} \):
\[ S_{n-1} = c(n-1)^2 = c(n^2 - 2n + 1) = cn^2 - 2cn + c \]

Now, substituting back into the formula for \( a_n \):
\[ a_n = S_n - S_{n-1} = cn^2 - (cn^2 - 2cn + c) = 2cn - c \]

### Step 2: Write the nth term in a simpler form
We can factor out \( c \):
\[ a_n = c(2n - 1) \]

### Step 3: Find the sum of squares of the first n terms
The sum of the squares of the first n terms is given by:
\[ \sum_{k=1}^{n} a_k^2 \]
Substituting \( a_k = c(2k - 1) \):
\[ \sum_{k=1}^{n} a_k^2 = \sum_{k=1}^{n} (c(2k - 1))^2 = c^2 \sum_{k=1}^{n} (2k - 1)^2 \]

### Step 4: Expand the summation
Now we need to compute \( \sum_{k=1}^{n} (2k - 1)^2 \):
\[ (2k - 1)^2 = 4k^2 - 4k + 1 \]
Thus,
\[ \sum_{k=1}^{n} (2k - 1)^2 = \sum_{k=1}^{n} (4k^2 - 4k + 1) = 4\sum_{k=1}^{n} k^2 - 4\sum_{k=1}^{n} k + \sum_{k=1}^{n} 1 \]

Using the formulas for the sums:
- \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)
- \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)

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

### Step 5: Simplify the expression
Calculating each term:
1. \( 4 \cdot \frac{n(n+1)(2n+1)}{6} = \frac{2n(n+1)(2n+1)}{3} \)
2. \( 4 \cdot \frac{n(n+1)}{2} = 2n(n+1) \)
3. \( \sum_{k=1}^{n} 1 = n \)

Combining these:
\[ \sum_{k=1}^{n} (2k - 1)^2 = \frac{2n(n+1)(2n+1)}{3} - 2n(n+1) + n \]

### Step 6: Final expression
Combining all terms gives:
\[ \sum_{k=1}^{n} (2k - 1)^2 = \frac{2n(n+1)(2n+1)}{3} - \frac{6n(n+1)}{3} + n = \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3} \]

### Step 7: Substitute back to find the sum of squares
Thus, the sum of squares of the first n terms is:
\[ \sum_{k=1}^{n} a_k^2 = c^2 \cdot \frac{2n(n+1)(2n+1) - 6n(n+1) + 3n}{3} \]

### Final Result
The final expression for the sum of squares of the first n terms of the AP is:
\[ \sum_{k=1}^{n} a_k^2 = \frac{c^2}{3} \left( 4n^3 + n \right) \]