Find the sum of the series
`3xx5+ 5xx7+ 7xx9+ ..` to n terms

Answer

To find the sum of the series \(3 \times 5 + 5 \times 7 + 7 \times 9 + \ldots\) up to \(n\) terms, we can follow these steps:

### Step 1: Identify the terms of the series
The series consists of terms that can be expressed as products of two sequences:
- The first sequence: \(3, 5, 7, \ldots\)
- The second sequence: \(5, 7, 9, \ldots\)

### Step 2: Determine the general term of the first sequence
The first sequence is an arithmetic progression (AP) where:
- First term \(a = 3\)
- Common difference \(d = 2\)

The \(n\)-th term of this AP can be expressed as:
\[ a_n = 3 + (n-1) \cdot 2 = 2n + 1 \]

### Step 3: Determine the general term of the second sequence
The second sequence is also an AP where:
- First term \(a = 5\)
- Common difference \(d = 2\)

The \(n\)-th term of this sequence can be expressed as:
\[ b_n = 5 + (n-1) \cdot 2 = 2n + 3 \]

### Step 4: Write the general term of the series
The general term of the series can be written as:
\[ T_n = a_n \cdot b_n = (2n + 1)(2n + 3) \]

### Step 5: Expand the general term
Expanding \(T_n\):
\[ T_n = (2n + 1)(2n + 3) = 4n^2 + 6n + 2n + 3 = 4n^2 + 8n + 3 \]

### Step 6: Find the sum of the series up to \(n\) terms
The sum \(S_n\) of the first \(n\) terms can be expressed as:
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (4k^2 + 8k + 3) \]

### Step 7: Break down the sum
We can separate the sum:
\[ S_n = 4\sum_{k=1}^{n} k^2 + 8\sum_{k=1}^{n} k + \sum_{k=1}^{n} 3 \]

### Step 8: Use formulas for 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 equation:
\[ S_n = 4 \cdot \frac{n(n+1)(2n+1)}{6} + 8 \cdot \frac{n(n+1)}{2} + 3n \]

### Step 9: Simplify the expression
Now, simplifying \(S_n\):
\[ S_n = \frac{2n(n+1)(2n+1)}{3} + 4n(n+1) + 3n \]

Finding a common denominator (which is 3):
\[ S_n = \frac{2n(n+1)(2n+1) + 12n(n+1) + 9n}{3} \]

### Step 10: Combine like terms
Combining the terms in the numerator:
\[ S_n = \frac{2n(n+1)(2n+1) + 12n(n+1) + 9n}{3} \]

### Final Result
Thus, the sum of the series up to \(n\) terms is:
\[ S_n = \frac{2n(n+1)(2n+1) + 12n(n+1) + 9n}{3} \]