If `1/(1!11!)+1/(3!9!)+1/(5!7!)=(2^n)/(m!)` then the value of m + n is

Options

A. 18
B. 23
C. 12
D. 22

Answer

To solve the equation

\[ \frac{1}{1! \cdot 11!} + \frac{1}{3! \cdot 9!} + \frac{1}{5! \cdot 7!} = \frac{2^n}{m!}, \]

we will follow these steps:

### Step 1: Rewrite the left-hand side
We notice that the denominators can be expressed in terms of \(12!\):
\[ \frac{1}{1! \cdot 11!} = \frac{1}{12!} \cdot \frac{12!}{1! \cdot 11!} = \frac{1}{12!} \cdot \binom{12}{1}, \]
\[ \frac{1}{3! \cdot 9!} = \frac{1}{12!} \cdot \frac{12!}{3! \cdot 9!} = \frac{1}{12!} \cdot \binom{12}{3}, \]
\[ \frac{1}{5! \cdot 7!} = \frac{1}{12!} \cdot \frac{12!}{5! \cdot 7!} = \frac{1}{12!} \cdot \binom{12}{5}. \]

### Step 2: Combine the fractions
Now we can combine these fractions:
\[ \frac{1}{12!} \left( \binom{12}{1} + \binom{12}{3} + \binom{12}{5} \right). \]

### Step 3: Use the binomial theorem
The sum of the binomial coefficients for odd indices can be derived from the binomial theorem. Specifically, we know:
\[ (1 + 1)^{12} = 2^{12} \quad \text{and} \quad (1 - 1)^{12} = 0. \]
From these, we can deduce:
\[ \text{Sum of odd indexed terms} = \frac{2^{12}}{2} = 2^{11}. \]

### Step 4: Substitute back into the equation
Thus, we have:
\[ \frac{1}{12!} \cdot 2^{11} = \frac{2^n}{m!}. \]

### Step 5: Compare both sides
From this equation, we can equate:
\[ \frac{2^{11}}{12!} = \frac{2^n}{m!}. \]
This implies:
- \(n = 11\)
- \(m! = 12!\)

### Step 6: Solve for \(m\)
Since \(m! = 12!\), we find that \(m = 12\).

### Step 7: Calculate \(m + n\)
Now, we can find \(m + n\):
\[ m + n = 12 + 11 = 23. \]

Thus, the final answer is:
\[ \boxed{23}. \]