If the sum of first n terms of an A.P. is `cn^(2)` then the sum of squares of these n terms is

Answer

To find the sum of the squares of the first n terms of an arithmetic progression (A.P.) given that the sum of the first n terms is \( S_n = cn^2 \), we can follow these steps:

### Step 1: Write the expression for \( S_n \)
The sum of the first n terms of an A.P. is given as:
\[ S_n = cn^2 \]

### Step 2: Write the expression for \( S_{n-1} \)
Now, let's express the sum of the first \( n-1 \) terms:
\[ S_{n-1} = c(n-1)^2 = c(n^2 - 2n + 1) = cn^2 - 2cn + c \]

### Step 3: Find the nth term \( T_n \)
The nth term \( T_n \) can be found using the relationship:
\[ T_n = S_n - S_{n-1} \]
Substituting the expressions we have:
\[ T_n = (cn^2) - (cn^2 - 2cn + c) = 2cn - c \]

### Step 4: Find the sum of squares of the first n terms
We need to find \( \sum_{k=1}^{n} T_k^2 \). We know:
\[ T_k = 2ck - c \]
Thus,
\[ T_k^2 = (2ck - c)^2 = 4c^2k^2 - 4c^2k + c^2 \]

### Step 5: Sum \( T_k^2 \) from 1 to n
Now, we will sum \( T_k^2 \):
\[ \sum_{k=1}^{n} T_k^2 = \sum_{k=1}^{n} (4c^2k^2 - 4c^2k + c^2) \]
This can be separated into three sums:
\[ = 4c^2 \sum_{k=1}^{n} k^2 - 4c^2 \sum_{k=1}^{n} k + \sum_{k=1}^{n} c^2 \]

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

We can substitute these into our expression:
\[ = 4c^2 \cdot \frac{n(n+1)(2n+1)}{6} - 4c^2 \cdot \frac{n(n+1)}{2} + c^2n \]

### Step 7: Simplify the expression
Now, we simplify:
\[ = \frac{4c^2n(n+1)(2n+1)}{6} - \frac{4c^2n(n+1) \cdot 3}{6} + c^2n \]
\[ = \frac{4c^2n(n+1)(2n+1 - 3)}{6} + c^2n \]
\[ = \frac{4c^2n(n+1)(2n-2)}{6} + c^2n \]
\[ = \frac{2c^2n(n+1)(n-1)}{3} + c^2n \]

### Final Result
The final expression for the sum of squares of the first n terms of the A.P. is:
\[ \sum_{k=1}^{n} T_k^2 = \frac{2c^2n(n+1)(n-1)}{3} + c^2n \]