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13

The denominator of each term is $$(n-2)!+(n-1)!+n!=(n-2)!(1+n-1+(n-1)n) = (n-2)!\,n^2,$$ so each term simplifies to $$\frac{n}{(n-2)!n^2}=\frac{1}{(n-2)!n}=\frac{n-1}{n!}=\frac{1}{(n-1)!}-\frac{1}{n!},$$ and now you can see that the series telescopes.


11

In the May 2013 Fibonacci Quarterly (Vol. 51, Num. 2) pages 163-173, Michael Hirschhorn proved this result: Let $P(n) =\prod\limits_{k=0}^n \binom{n}{k} $. Then (this is going to be a pain to enter), as $n \to \infty$, $$P(n) \sim C^{-1}\dfrac{e^{n(n+2)/2}}{n^{(3n+2))/6}(2\pi)^{(2n+1)/4}} \exp\left(-\sum\limits_{p \ge ...


10

Just dividing by $2,3,4,\ldots$ till you get down to $1$ works perfectly, but we can be smarter (if we know that the equation has an integer solution). Start by locating the largest prime number present in the factorisation of $n!$: this is $7$. Because of the uniqueness of prime factorisation we now know that $n$ must be smaller than the next prime number ...


7

A posibble form is: $$\prod_{i=1}^ni!$$ In other way is $$1^n\cdot2^{n-1}\cdot3^{n-2}\cdot\cdots\cdot(n-1)^2\cdot n^1$$


5

Algebra: note that $\;r!=r(r-1)!\;,\;\;\frac{r!}r=(r-1)!\;$ , and also correcting your equalities we have $$\binom n{k-1}=\frac{n!}{(k-1)!(n-k+1)!}\;,\;\;\binom nk=\frac{n!}{k!(n-k)!}$$ so that we actually should have on the RHS of your last equality $$\frac{n!}{(k-1)!\cdot(n-k+1)!} + \frac{n!}{k!\cdot(n-k)!}=n!\frac{k+n-k+1}{k!(n-k+1)!}=$$ ...


4

I can give you an other way of proving this without any calculus. What is ${{n}\choose{k}}$? This is the number of ways to choose k elements among n ones, without specifying an order. So let's have the set: A=$(a_1...a_n)$ n elements ( tennis balls per say). Let's look at one particularly: $a_i$ because it is the only red ball of the set. The number of ...


4

Start writing all the numbers $1,2,..,n$ using the factors in the expression, until you run out: You have a $2$, then you are left with $2^73^4 5^2 7$ Then you have , so far , $2!$ Now, use a $3$ , and you are left with $2^73^3 5^2 7$ And you have $3! =2.3 $ Now you want to construct the $4$ , so use two copies of $2$ , and you have $2^53^3 5^2 7$ In ...


4

$p$ divides $n!$. So it cannot divide $n!+1$.


3

You have $n$ choices for the first element in the list, $n-1$ for the second, ..., $n-(k-1)$ for the $k$-th. So the number is $$ n \cdot (n-1) \cdots (n-(k-1)) = n \cdot (n-1) \cdots (n-(k-1)) \frac{(n-k)!}{(n-k)!} = \frac{n!}{(n-k)!} $$


3

In Factorial Inequality problem it is shown that $$n!<\left(\frac n2\right)^n.$$ For $n=2000$ we have $$2000!<\left(\frac {2000}{2}\right)^{2000}=1000^{2000}.$$


3

$$\sin x=\frac{x^1}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\frac{x^8}{8!}-\frac{x^9}{9!}+\cdots$$ $$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\frac{x^8}{8!}-\frac{x^{10}}{10!}+\cdots$$ $$e^x=1+\frac{x^1}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^5}{5!}+\cdots$$ The sine and cosine functions are important in trigonometry, ...


2

During a mathematical education program you will usually encounter it in calculus, for example Taylor's theorem $$ f(x) = \sum_{k=0}^\infty \frac{f^{(n)}(x_0)}{k!}(x-x_0)^k. $$ and the binomial theorem $$ (a+b)^n = \sum_{k=0}^n \binom{n}{k} a^k b^{n-k}, \quad \binom{n}{k} = \frac{n!}{k! (n-k)!} $$ or combinatorics (art of counting). Permutations show up ...


2

Converting above realization to an answer. We prove $ \left( n! \right)^2 \ge n^n $ by means of induction. For $n=1$, we have $ 1 \ge 1 $, which is true. Now suppose that $ \left( n! \right)^2 \ge n^n $. Then, $$ \begin {align*} \left( (n+1)! \right)^2 &= \left( n + 1 \right)^2 \left( n! \right)^2 \\&\ge \left( n + 1 \right)^2 n^n. \end ...


2

you need to add only $1!+2!+3!+4!$ which results in last digit as $3$, since number more than $5!$ or $5!$ itself, will have last digit as $0$, due to having a factor $10$.


2

Assuming that $a$ and $b$ are natural numbers: $$ \frac{(a+b-1)!}{(b-1)!}=b(1+b)(2+b)\cdot\ldots\cdot(a-1+b). $$


2

There are many functions $F\colon[0,\infty)$ such that $f(n)=n!$ for $n=0,1,2\dots$. How to choose the right one? It seems reasonable to require that $F$ be continuous, but this is not enough. An important property of the factorial is $$ (n+1)!=(n+1)\,n!. $$ We then add the requirement $$ F(x+1)=(x+1)\,F(x),\quad x>0. $$ There are still many functions ...


2

It is sometimes referred to as the superfactorial of n (although there is at least one other function that's also called superfactorial), and its values are given by OEIS sequence A000178. I don't believe there are any shortcuts to computing it, other than noting that $sf(n) = \prod_{1 \le i \lt j \le n}\left( j - i \right)$


2

Factorials are bigger than Bell numbers, except for the initial cases when there is equality. A comment from Emeric Deutsch on OEIS A048742 says that the difference counts Number of permutations of $[n]$ which have at least one cycle that has at least one inversion when written with its smallest element in the first position. Example: $a(4)=9$ ...


2

The definition of factorial is $n!=n\times(n-1)\times\dots\times1$, or if you prefer $$\begin{eqnarray} 0! &=& 1\\ n! &=&n\times(n-1)!\end{eqnarray}$$ Thus, if $p<q$, you have that $p!$ divides $q!$, so $GCD(p!,q!)=p!$, and $LCM(p!,q!)=q!$.


2

By definition, $$e=\lim_{n\to\infty}\left(1+\frac1n\right)^n.$$ Using the binomial theorem, the $k^{th}$ term of the development is $${\binom nk}\frac1{n^k}=\frac{n(n-1)(n-2)\dots(n-k+1)}{k!.n.n.n\dots n},$$ and $$\lim_{n\to\infty}{\binom nk}\frac1{n^k}=\frac1{k!}.$$ For example, ...


2

These two familiar sums are the Taylor series for $e^x$ about $0$. To get $e$ itself, you evaluate this series at $x=1.$ Derivation: The $n$th term of the Taylor series of a function $f$ about $a$ is $$ \frac{f^{(n)}(a)}{n!} (x-a)^n.$$ But if $f(x) \triangleq e^x$, then $f'(x) = e^x$, and by an inductive argument, $f^{(n)}(x) = e^x$ for every positive ...


2

The formula, $$\sum_{k=0}^\infty\frac{2^{-5k}(6k+1)((2k-1)!!)^3}{4(k!)^3} = \frac{1}{\pi}\tag1$$ or its equivalent form, $$\sum_{k=0}^\infty \frac{(2k)!^3}{k!^6}\frac{6k+1}{(2^8)^{k+1/2}} = \frac{4}{\pi}\tag2$$ and the similar, $$\sum_{k=0}^\infty (-1)^k \frac{(2k)!^3}{k!^6}\frac{6k+1}{(2^9)^{k+1/2}} = \frac{8}{\pi}\tag3$$ belong to the Ramanujan-type ...


2

EDIT: Wrote a command to do sum of distinct factorials by greedy algorithm; here are the perfect powers I have, not allowing $0!$. jagy@phobeusjunior:~$ ./greedy_factorial Sat Dec 13 12:25:29 PST 2014 0 1 1 8 3 2 9 3 2 1 25 4 1 27 4 2 1 32 4 3 2 121 5 1 128 5 3 2 144 5 4 729 6 3 2 1 ...


2

The ratio of two consecutive terms is $$\frac{x_{n+1}}{x_n}=\frac{2n+1}{2n+2}=1+\frac{\color{red}{a}}n+o\left(\frac1n\right),$$ with $$\color{red}{a}=-\frac12.$$ Note that $\color{red}{a}\lt0$ hence the sequence $(x_n)$ is eventually decreasing and its limit is $0$. Furthermore (but this is not needed for the limit), at least in a loose sense of $\approx$, ...


2

Consider the $n!$ permutationsof $n$ elements. They are all different. But you only care about the first $k$ elements, so the permutations that differ only in the order of the last $n-k$ elements (there are $(n-k)!$ of them) must be counted once only. ab|cd ab|dc ac|bd ac|db ad|bc ad|cb ba|cd ba|dc bc|ad bc|da bd|ac bd|ca ca|bd ...


2

It is well known that $$ \binom{n}{k}=\frac{n!}{(n-k)!k!} $$ gives the number of ways to choose $k$ items from a list of $n$ items. Now this does not take order into account. For example, if you have the numbers $1$, $2$ and $3$, and you want to choose two numbers between them, you will have $$ 1,2\\ 1,3\\ 2,3 $$ as possibilities. That's $\frac{3!}{1!2!}=3$ ...


2

The difference sequence of any polynomial $Cx^n + \cdots$ (where here and elsewhere, $\cdots$ hides terms of lower degree) is of the form $Cnx^{n-1} + \cdots$, since $(x+1)^n - x^n = nx^{n-1} + \cdots$ (just open the binomial). If we start with $x^n + \cdots$ then we get $nx^{n-1} + \cdots$, then $n(n-1)x^{n-2} + \cdots$, then $n(n-1)(n-2)x^{n-3} + \cdots$, ...


2

Hint: Since the panels are square and you have to cover everything, you want the largest integer that divides 280, 336, and 168 simultaneously, i.e. you want the greatest common divisor.


1

Hint: for simplicity, write the powers of 2 and 3 in integers: $2 = 2^1$ $3 = 3^1$ $4 = 2^2$ $5 = x$ $6 = 2.3$ $7 = x'$ $8 = 2^3$ $9 = 3^2$ $10 = 2.x''$ Add the prime factors' exponents and you see that $10! = 2^8 .3^4 .x'''$. If you went up to the factorial of $12=2^2.3$ you see that you would have too many powers of $2$ and $3$, so the answer is 10 or ...


1

Let's first try to find a nice bound. If $n$ is at least 11, then $n!$ will be divisible by $11$ because $11$ will be one of $n!$'s factors. Hence $n$ is at most $10$. However, if $n$ is less than $10$, then in $1$, $2$, ..., $n$, there will be at most one multiple of $5$ ($10$ is the second multiple of $5$). Thus $5^2$ cannot be a factor of $n!$ if ...



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