Let `a_(n)=(827^(n))/(n!)` for `n in N` , then `a_(n)` is greatest when

Options

A. `n=824`
B. `n= 825`
C. n = 826
D. n = 828

Answer

To find the value of \( n \) for which \( a_n = \frac{827^n}{n!} \) is greatest, we can analyze the ratio of consecutive terms \( a_{n+1} \) and \( a_n \).

### Step-by-step Solution:

1. **Define the terms**:
\[ a_n = \frac{827^n}{n!} \]
\[ a_{n+1} = \frac{827^{n+1}}{(n+1)!} \]

2. **Form the ratio \( \frac{a_{n+1}}{a_n} \)**:
\[ \frac{a_{n+1}}{a_n} = \frac{827^{n+1}}{(n+1)!} \cdot \frac{n!}{827^n} = \frac{827 \cdot 827^n}{(n+1) \cdot n!} \cdot \frac{n!}{827^n} = \frac{827}{n+1} \]

3. **Set up the inequality**:
To find when \( a_n \) is greatest, we need to find when \( \frac{a_{n+1}}{a_n} \) is greater than or equal to 1:
\[ \frac{827}{n+1} \geq 1 \]

4. **Solve the inequality**:
\[ 827 \geq n + 1 \]
\[ n \leq 826 \]

5. **Conclusion**:
The maximum value of \( n \) for which \( a_n \) is greatest is \( n = 826 \).

### Final Answer:
The value of \( n \) for which \( a_n \) is greatest is \( n = 826 \).