If ` a,b`and `c` are three consecutive positive integers such that `1/(a!)+1/(b!)=lambda/(c!)` then find the value of root under lambda .

Options

A. a
B. b
C. c
D. a+b+c

Answer

To solve the problem, we need to find the value of \(\sqrt{\lambda}\) given that \(a\), \(b\), and \(c\) are three consecutive positive integers such that:

\[ \frac{1}{a!} + \frac{1}{b!} = \frac{\lambda}{c!} \]

### Step-by-Step Solution:

1. **Define the Consecutive Integers**:
Since \(a\), \(b\), and \(c\) are consecutive integers, we can express them in terms of \(b\):
- Let \(b = n\)
- Then \(a = n - 1\)
- And \(c = n + 1\)

2. **Substituting into the Equation**:
Substitute \(a\), \(b\), and \(c\) into the given equation:
\[ \frac{1}{(n-1)!} + \frac{1}{n!} = \frac{\lambda}{(n+1)!} \]

3. **Finding a Common Denominator**:
The left-hand side can be simplified:
\[ \frac{1}{(n-1)!} + \frac{1}{n!} = \frac{n + 1}{n!} \]
This is because:
\[ \frac{1}{(n-1)!} = \frac{n}{n!} \quad \text{and} \quad \frac{1}{n!} = \frac{1}{n!} \]
So, adding these gives:
\[ \frac{n + 1}{n!} \]

4. **Setting the Equation**:
Now we have:
\[ \frac{n + 1}{n!} = \frac{\lambda}{(n + 1)!} \]

5. **Cross-Multiplying**:
Cross-multiply to eliminate the fractions:
\[ (n + 1)(n + 1)! = \lambda \cdot n! \]
Simplifying this gives:
\[ (n + 1) \cdot (n + 1) \cdot n! = \lambda \cdot n! \]
Thus, we can cancel \(n!\) from both sides (assuming \(n! \neq 0\)):
\[ (n + 1)^2 = \lambda \]

6. **Finding \(\sqrt{\lambda}\)**:
Now we need to find \(\sqrt{\lambda}\):
\[ \sqrt{\lambda} = \sqrt{(n + 1)^2} = n + 1 \]

7. **Relating Back to \(c\)**:
Since \(c = n + 1\), we can conclude:
\[ \sqrt{\lambda} = c \]

### Final Answer:
Thus, the value of \(\sqrt{\lambda}\) is:
\[ \sqrt{\lambda} = c \]