Question

Let ${a_n}$ $ = \dfrac{{{{(1000)}^n}}}{{n!}}$ for $n \in N$, then ${a_n}$ is greatest, when
A. $n = 998$
B. $n = 999$ and $n = 1000$
C. $n = 1002$ and $n = 1001$
D. none of these

Answer 1

Hint: First, find the value of $1^{st}$ term and $2^{nd}$ term to check whether the function is increasing or decreasing. After that find the nth term until the function is increasing. The term before the $999^{th}$ term is less than the preceding term. Now, check whether the $1000^{th}$ term is increasing or decreasing.

Complete step-by-step answer:
Given: - ${a_n}$ $ = \dfrac{{{{(1000)}^n}}}{{n!}}$ for $n \in N$
The first term is 1000, the second term is 500000, so the sequence starts out increasing. Let's investigate to see how many terms the sequence is increasing.
The (n+1)th term will be,
${a_{n + 1}} = \dfrac{{{{\left( {1000} \right)}^n}}}{{n + 1}}$
Now, divide $(n+1)^{th}$ term by $n^{th}$ term, we get,
$\dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{{\dfrac{{{{\left( {1000} \right)}^{n + 1}}}}{{\left( {n + 1} \right)!}}}}{{\dfrac{{{{\left( {1000} \right)}^n}}}{{n!}}}}$
Multiply the ${a_n}$ term in reverse with ${a_{n + 1}}$ the term,
\[\dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{{{{1000}^{n + 1}}}}{{\left( {n + 1} \right)!}} \times \dfrac{{n!}}{{{{1000}^n}}}\]
Now expand the terms,
 \[\dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{{1000 \times {{1000}^n}}}{{\left( {n + 1} \right) \times n!}} \times \dfrac{{n!}}{{{{1000}^n}}}\]
Cancel out the common factors from both numerator and denominator,
$\dfrac{{{a_{n + 1}}}}{{{a_n}}} = \dfrac{{1000}}{{n + 1}}$
For, n = 1, 2, 3, ………, 999
$\dfrac{{{a_{n + 1}}}}{{{a_n}}} \geqslant 1$
For n= 1000,
${a_{1000}} = \dfrac{{{{1000}^{1000}}}}{{1000!}}$
Take one multiple from both numerator and denominator,
${a_{1000}} = \dfrac{{1000 \times {{1000}^{999}}}}{{1000 \times 999!}}$
Cancel out the common factor from
$ = \dfrac{{{{1000}^{999}}}}{{999!}}$
The term is equal to the $999^{st}$ term,
${a_{1000}} = {a_{999}}$
So, the 999th term is the last term for which the sequence is increasing.
After the $1000^{st}$ term, the multiplication will be by a fraction less than 1, beginning with multiplying by $\dfrac{{1000}}{{1001}}$ and so the sequence will begin to decrease with the $1001^{st}$ term.
So, the 999th and the 1000th terms, which are equal, are the largest terms of the sequence.
Thus, the sequence attains a maximum at exactly two values of n, 999, and 1000.

Hence, option (b) is the correct answer.

Note: The students might make mistakes when finding the term until the value is increasing.
The factorial of a positive integer n, denoted by n! is the product of all positive integers less than or equal to n.
The factorial of 0 is also 1.