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 determine when the sequence \( a_n = \frac{827^n}{n!} \) is greatest, we can analyze the ratio of consecutive terms, \( \frac{a_{n+1}}{a_n} \).

### Step-by-Step Solution:

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

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

3. **Determine when the sequence is increasing**:
The sequence \( a_n \) is increasing when \( \frac{a_{n+1}}{a_n} > 1 \):
\[ \frac{827}{n+1} > 1 \]
This simplifies to:
\[ 827 > n + 1 \quad \Rightarrow \quad n < 826 \]

4. **Determine when the sequence is decreasing**:
The sequence \( a_n \) starts to decrease when \( \frac{a_{n+1}}{a_n} < 1 \):
\[ \frac{827}{n+1} < 1 \]
This simplifies to:
\[ 827 < n + 1 \quad \Rightarrow \quad n > 826 \]

5. **Conclusion**:
The sequence \( a_n \) increases for \( n < 826 \) and decreases for \( n > 826 \). Therefore, the maximum value occurs at \( n = 826 \).

Thus, the value of \( n \) for which \( a_n \) is greatest is:
\[ \boxed{826} \]