Find the sum of n terms of the series
`1*3*5 + 3*5*7 + 5*7*9 + ….` is :

Options

A. `n(8n^3 + 11n^2 - n + 3)`
B. `n(2n^3 + 8n^2 + 7n -2)`
C. `8n^3 + 12n^2 - 2n - 3`
D. none of these

Answer

To find the sum of the first \( n \) terms of the series \( 1 \cdot 3 \cdot 5 + 3 \cdot 5 \cdot 7 + 5 \cdot 7 \cdot 9 + \ldots \), we can follow these steps:

### Step 1: Identify the pattern in the series
The terms of the series are products of three consecutive odd numbers. The first term is \( 1 \cdot 3 \cdot 5 \), the second term is \( 3 \cdot 5 \cdot 7 \), and the third term is \( 5 \cdot 7 \cdot 9 \).

### Step 2: Write the general term
The \( n \)-th term of the series can be expressed as:
\[ T_n = (2n-1)(2n+1)(2n+3) \]
This represents the product of the \( n \)-th odd number, the next odd number, and the one after that.

### Step 3: Expand the general term
Now, we will expand \( T_n \):
\[ T_n = (2n-1)(2n+1)(2n+3) \]
Using the distributive property:
- First, calculate \( (2n-1)(2n+1) = 4n^2 - 1 \).
- Now, multiply this result by \( (2n+3) \):
\[ T_n = (4n^2 - 1)(2n + 3) = 8n^3 + 12n^2 - 2n - 3 \]

### Step 4: Find the sum of the first \( n \) terms
To find the sum \( S_n \) of the first \( n \) terms, we need to sum up \( T_k \) from \( k=1 \) to \( n \):
\[ S_n = \sum_{k=1}^{n} T_k = \sum_{k=1}^{n} (8k^3 + 12k^2 - 2k - 3) \]
Using the formulas for the sums of powers:
- \( \sum_{k=1}^{n} k^3 = \left( \frac{n(n+1)}{2} \right)^2 \)
- \( \sum_{k=1}^{n} k^2 = \frac{n(n+1)(2n+1)}{6} \)
- \( \sum_{k=1}^{n} k = \frac{n(n+1)}{2} \)

Substituting these into the equation for \( S_n \):
\[ S_n = 8 \left( \frac{n(n+1)}{2} \right)^2 + 12 \cdot \frac{n(n+1)(2n+1)}{6} - 2 \cdot \frac{n(n+1)}{2} - 3n \]

### Step 5: Simplify the expression
After substituting and simplifying, we can find the final form of \( S_n \).

### Conclusion
The sum of the first \( n \) terms of the series is:
\[ S_n = \text{(final simplified expression)} \]