Find the sum of n terms of the series `1^2 + 4^2+ 7^2 + .........`

Explanation

To find the sum of the first \( n \) terms of the series \( 1^2 + 4^2 + 7^2 + \ldots \), we can follow these steps:

### Step 1: Identify the sequence
The terms of the series are \( 1, 4, 7, \ldots \). This is an arithmetic progression (AP) where:
- The first term \( a = 1 \)
- The common difference \( d = 3 \)

### Step 2: Find the \( n \)-th term of the AP
The \( n \)-th term of an AP can be calculated using the formula:
\[ T_n = a + (n - 1)d \]
Substituting the values of \( a \) and \( d \):
\[ T_n = 1 + (n - 1) \cdot 3 = 3n - 2 \]

### Step 3: Write the sum of squares of the terms
We need to find the sum of the squares of the first \( n \) terms:
\[ S_n = T_1^2 + T_2^2 + T_3^2 + \ldots + T_n^2 = (3 \cdot 1 - 2)^2 + (3 \cdot 2 - 2)^2 + (3 \cdot 3 - 2)^2 + \ldots + (3n - 2)^2 \]
This can be expressed as:
\[ S_n = \sum_{k=1}^{n} (3k - 2)^2 \]

### Step 4: Expand the square
Now we expand \( (3k - 2)^2 \):
\[ (3k - 2)^2 = 9k^2 - 12k + 4 \]
Thus, we can rewrite the sum:
\[ S_n = \sum_{k=1}^{n} (9k^2 - 12k + 4) \]
This can be separated into three sums:
\[ S_n = 9\sum_{k=1}^{n} k^2 - 12\sum_{k=1}^{n} k + \sum_{k=1}^{n} 4 \]

### Step 5: Use formulas for the sums
We use the formulas for the sums:
1. The sum of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k = \frac{n(n + 1)}{2} \]
2. The sum of the squares of the first \( n \) natural numbers:
\[ \sum_{k=1}^{n} k^2 = \frac{n(n + 1)(2n + 1)}{6} \]

Substituting these formulas into our expression for \( S_n \):
\[ S_n = 9 \cdot \frac{n(n + 1)(2n + 1)}{6} - 12 \cdot \frac{n(n + 1)}{2} + 4n \]

### Step 6: Simplify the expression
Now we simplify each term:
1. The first term becomes:
\[ \frac{9n(n + 1)(2n + 1)}{6} = \frac{3n(n + 1)(2n + 1)}{2} \]
2. The second term becomes:
\[ 12 \cdot \frac{n(n + 1)}{2} = 6n(n + 1) \]
3. The third term is simply \( 4n \).

Combining these:
\[ S_n = \frac{3n(n + 1)(2n + 1)}{2} - 6n(n + 1) + 4n \]

### Step 7: Combine like terms
To simplify further, we can factor out \( n(n + 1) \):
\[ S_n = n(n + 1) \left( \frac{3(2n + 1)}{2} - 6 \right) + 4n \]
This simplifies to:
\[ S_n = n(n + 1) \left( \frac{3(2n + 1) - 12}{2} \right) + 4n \]

### Final Step: Write the final expression
After simplifying, we can express \( S_n \) in a more compact form. The final result will be:
\[ S_n = \frac{n(n + 1)(3n - 5)}{2} \]