The sum of n terms of an A.P. is 4n(n - 1), then the sum of their squares is

Options

A. `n^(3)`
B. `(32)/(3) n (n - 1) (2n - 1)`
C. `(16)/(3) n (n - 1) (2n - 1)`
D. `4n (n - 1) (2n - 1)`

Answer

To solve the problem, we need to find the sum of the squares of the terms of an arithmetic progression (A.P.) given that the sum of the first n terms \( S_n \) is \( 4n(n - 1) \).

### Step-by-step Solution:

1. **Identify the sum of n terms**:
\[ S_n = 4n(n - 1) \]

2. **Find the nth term \( t_n \)**:
The nth term can be found using the formula:
\[ t_n = S_n - S_{n-1} \]
We first need to compute \( S_{n-1} \):
\[ S_{n-1} = 4(n-1)((n-1) - 1) = 4(n-1)(n-2) = 4(n^2 - 3n + 2) \]
Now, substituting \( S_n \) and \( S_{n-1} \):
\[ t_n = S_n - S_{n-1} = 4n(n - 1) - 4(n^2 - 3n + 2) \]
Simplifying this:
\[ t_n = 4n(n - 1) - 4(n^2 - 3n + 2) = 4n^2 - 4n - 4n^2 + 12n - 8 = 8n - 8 \]
Thus,
\[ t_n = 8(n - 1) \]

3. **Find the sum of the squares of the terms**:
We need to find \( \sum_{k=1}^{n} t_k^2 \):
\[ t_k = 8(k - 1) \implies t_k^2 = (8(k - 1))^2 = 64(k - 1)^2 \]
Therefore,
\[ \sum_{k=1}^{n} t_k^2 = \sum_{k=1}^{n} 64(k - 1)^2 = 64 \sum_{k=1}^{n} (k - 1)^2 \]

4. **Calculate \( \sum_{k=1}^{n} (k - 1)^2 \)**:
We know that:
\[ \sum_{k=1}^{n} (k - 1)^2 = \sum_{j=0}^{n-1} j^2 = \frac{(n-1)n(2(n-1)+1)}{6} \]
Substituting this back:
\[ \sum_{k=1}^{n} (k - 1)^2 = \frac{(n-1)n(2n-1)}{6} \]

5. **Combine the results**:
Thus, the sum of the squares becomes:
\[ \sum_{k=1}^{n} t_k^2 = 64 \cdot \frac{(n-1)n(2n-1)}{6} = \frac{64(n-1)n(2n-1)}{6} \]

### Final Answer:
The sum of the squares of the terms of the A.P. is:
\[ \frac{64(n-1)n(2n-1)}{6} \]