If the sum to n terms of the series `(1)/(1*3*5*7)+(1)/(3*5*7*9)+(1)/(5*7*9*11)+"......"` is `(1)/(90)-(lambda)/(f(n))`, then find `f(0), f(1)` and `f(lambda)`

Options

A. `f(0)=15`
B. `f(1)=105`
C. `f(lambda)=(640)/(27)`
D. `lambda=(1)/(3)`

Explanation

To solve the problem, we need to find the values of \( f(0) \), \( f(1) \), and \( f(\lambda) \) based on the given series and its sum representation.

### Step 1: Identify the General Term of the Series

The series given is:
\[ \frac{1}{1 \cdot 3 \cdot 5 \cdot 7} + \frac{1}{3 \cdot 5 \cdot 7 \cdot 9} + \frac{1}{5 \cdot 7 \cdot 9 \cdot 11} + \ldots \]

The general term can be expressed as:
\[ T_n = \frac{1}{(2n-1)(2n+1)(2n+3)(2n+5)} \]

### Step 2: Simplify the General Term

We can rewrite the general term using partial fractions. The term can be manipulated as follows:
\[ T_n = \frac{1}{6} \left( \frac{1}{(2n-1)(2n+1)(2n+3)} - \frac{1}{(2n+1)(2n+3)(2n+5)} \right) \]

### Step 3: Sum the Series

The sum of the first \( n \) terms can be expressed as:
\[ S_n = \sum_{k=0}^{n-1} T_k \]
This results in a telescoping series where most terms cancel out. The remaining terms will be:
\[ S_n = \frac{1}{6} \left( \frac{1}{1 \cdot 3 \cdot 5} - \frac{1}{(2n+1)(2n+3)(2n+5)} \right) \]

### Step 4: Express the Sum in the Given Form

We know from the problem statement that:
\[ S_n = \frac{1}{90} - \frac{\lambda}{f(n)} \]

From our derived expression for \( S_n \):
\[ S_n = \frac{1}{6} \left( \frac{1}{15} - \frac{1}{(2n+1)(2n+3)(2n+5)} \right) \]

### Step 5: Equate and Identify \( f(n) \) and \( \lambda \)

By simplifying \( S_n \) further, we find:
\[ S_n = \frac{1}{90} - \frac{1}{6(2n+1)(2n+3)(2n+5)} \]

From this, we can identify:
\[ f(n) = (2n+1)(2n+3)(2n+5) \]
and
\[ \lambda = \frac{1}{6} \]

### Step 6: Calculate \( f(0) \), \( f(1) \), and \( f(\lambda) \)

1. **Calculate \( f(0) \)**:
\[ f(0) = (2 \cdot 0 + 1)(2 \cdot 0 + 3)(2 \cdot 0 + 5) = 1 \cdot 3 \cdot 5 = 15 \]

2. **Calculate \( f(1) \)**:
\[ f(1) = (2 \cdot 1 + 1)(2 \cdot 1 + 3)(2 \cdot 1 + 5) = 3 \cdot 5 \cdot 7 = 105 \]

3. **Calculate \( f(\lambda) \)** where \( \lambda = \frac{1}{6} \):
\[ f\left(\frac{1}{6}\right) = \left(2 \cdot \frac{1}{6} + 1\right)\left(2 \cdot \frac{1}{6} + 3\right)\left(2 \cdot \frac{1}{6} + 5\right) \]
This simplifies to:
\[ f\left(\frac{1}{6}\right) = \left(\frac{1}{3} + 1\right)\left(\frac{1}{3} + 3\right)\left(\frac{1}{3} + 5\right) = \left(\frac{4}{3}\right)\left(\frac{10}{3}\right)\left(\frac{16}{3}\right) = \frac{640}{27} \]

### Final Results

Thus, the values are:
- \( f(0) = 15 \)
- \( f(1) = 105 \)
- \( f(\lambda) = \frac{640}{27} \)