Prove that $\frac{(3n)!}{(3!)^n}$ is an integer where $n$ is a non-negative integer.  I can prove it with mathematical induction. Is there any other method?
120942
This is also a question on my exam paper that i proved by using mathematical induction. However, my tutor tells me that it can be proved without using mathematical induction. I really want to know how to deal with in another way.  Show that for any positive integer n, $(3n)!/(3!)^n$ is an integer.
37322
Prove using induction that $\forall n\in\mathbb N, \forall x\in \mathbb R: 0&lt;x&lt;1: (1-x)^n&lt;\frac 1 {1+nx}$   My attempt:   Base: for $n=1: 1-x&lt;\frac 1 {1+x}\iff 1-x^2&lt;1$, true since $0&lt;x&lt;1$.  Suppose the statement is true for $n$, prove for $n+1$:  $(1-x)^{n+1}=(1-x)(1-x)^{n}\overset{i.h}&lt;\frac{(1-x)}{1+nx}$  Now I got stuck, maybe another induction to show that $1+nx+x&lt;1+nx$? Is there another way?  Moreover, I was told it's wrong to begin with $(1-x)^{n+1}$ and reach to $\frac 1 {1+(n+1)x}$ but why? Is it assuming what I need to prove?
69970
The main question is :  Prove that $n^n&gt;1\times3\times5\times7\times\dots\times(2n-1)$  My approach :  We can write the R.H.S as,  $$\frac{(2n-1)!}{2\times4\times6\times\dots\times2(n-1)}$$ We can write $(2n-1)!$ as $(2n-3)!(2n-1)2(n-1)$  Thus, in this manner we can cancel out even terms till $n$ appears. I'm having trouble doing this. What I intend to do is say that any positive number's factorial is less than the number raised to itself. I welcome any alternate method. Please help.  P.S. I am still in high school and do not understand binomial theorem and induction.
64652
Is there any difference in proving "for any integer $n &gt; 0$, $3|(5^{2n}-1)$" and proving "for any integer $n \ge 0$, $3|(5^{2n} -1)$"? Or are they the same?
121445
Prove the following:  $$f_3+f_6+...f_{3n}= \frac 12(f_{3n+2}-1) \\ $$  For $n \ge 2$  Well I got the basis out of the way, so now I need to use induction: So that $P(k) \rightarrow P(k+1)$ for some integer $k \ge 2$  So, here are my first steps:  $$ \begin{align} &amp; \frac 12(f_{3k+2}-1) + f_{3k+3} = \\     &amp; = \frac 12(f_{3k+2}-1) + f_{3k+1} + f_{3k} +  f_{3k+1} \\    &amp; = \frac 12(f_{3k+2}-1) + f_{3k+1} + \frac 12(f_{3k+2}-1) +  f_{3k+1} \\    &amp; = f_{3k+2}-1 + 2 \cdot f_{3k+1}     \end{align} $$  And the fun stops around here. I don't see how to get to the conclusion: $\frac 12(f_{3k+5}-1) \\ $. Any help from this point would be great.
88080
I would like to prove by induction the following inequality:  $\frac{4^n}{n+1} &lt; \binom{2n}{n}$, for all natural numbers n > 1.  Any hints?
52765
I would like a proof that $$ {{n}\choose{k}} = \frac{n!}{k!(n-k)!} = 2^m $$ for $n,k,m\in \mathbb{N}$, only if $k=1$ or $k=n-1$.  It seems to me that this must be true since for other values of $k$ the numerator contains more factors that are not powers of 2 than the denominator. Furthermore, the numerator also contains larger factors than the denominator and thus can't all be cancelled. Nevertheless, I have been unable to form an elegant, watertight proof.
2770
Show that if $n&gt;2$, then $(n!)^2&gt;n^n$.     My work: I tried to apply induction. So, at the induction step, I need to prove, $n^n&gt;(n+1)^{n-1}$ Here, I tried to use induction again without any luck. I also took log of both sides, but I did not get anything. Please help!
88141
So I have to prove:      For each natural number greater than or equal to 3,   $$(1+\frac1n)^n&lt;n$$   My work: Basis step: $n=3$ $$\left(1+\frac13\right)^3&lt;3$$ $$\left(\frac43\right)^3&lt;3$$ $$\left(\frac{64}{27}\right)&lt;3$$ which is true.  Now the inductive step, assume $P(k)=\left(1+\frac1k\right)^k&lt;k$ to be true and prove $P(k+1)=\left(1+\frac1{k+1}\right)^{k+1}&lt;k+1$.   This is where I am stuck because usually you add or multiply by $k+1$ or some similar term.
15333
How to prove that $ 2&lt;(1+\frac{1}{n})^{n}$ for every integer $n&gt;1$. I was thinking by induction, it works for $n=2$ but then I couldn't move forward.
74594
Using permutation or otherwise, prove that $\displaystyle \frac{(n^2)!}{(n!)^n}$ is an integer,where $n$ is a positive integer.  I have no idea how to prove this..!!I am not able to even start this Can u give some hints or the solution.!cheers.!!
69440
Show by induction that for all natural numbers n>3  $(1+\frac{1}{n})^n&lt;n$  Let $(1+\frac{1}{n})^n&lt;n$ be true !   We show that $(1+\frac{1}{n+1})^{n+1})&lt;n+1$  $(1+\frac{1}{n})^n(1+\frac{1}{n})&lt;n(1+\frac{1}{n})$   $(1+\frac{1}{n})^{n+1}&lt;n+1$  So now we just need to show that   $(1+\frac{1}{n+1})^{n+1}&lt;(1+\frac{1}{n})^{n+1}$  So here , If I show that :   $(1+\frac{1}{n+1})&lt;(1+\frac{1}{n})$   Does that mean im done with exercise ?
81479
Show $$\frac{(2n)!}{n!\cdot 2^n}$$ is an integer for $n$ greater than or equal to $0$.   Could anyone please help me with this proving? Thanks!
120175
I want to justificate following equation:  $$\sum_{k=0}^n \frac{(-1)^k}{k!(n-k)!}\frac{1}{2k+1} = \frac{2^n}{(2n+1)!!}$$  I calculated the both sides for $n$ from 1 to 10 and it was true. How the mathematical induction can be applied to this equation? Or is there other way to justificate it?
63335
Prove that $1&lt;\dfrac{1}{n+1}+\dfrac{1}{n+2}+...+\dfrac{1}{3n+1}$.  By using the Mathematical induction. Suppose the statement holds for $n=k$.  Then for $n=k+1$. We have $\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}+\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}=(\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1})+(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$  we know $\dfrac{1}{k+1}+\dfrac{1}{k+2}+\dfrac{1}{k+3}+...+\dfrac{1}{3k+1}&gt;1$  What can we do for $(\dfrac{1}{3k+2}+\dfrac{1}{3k+3}+\dfrac{1}{3k+4}-\dfrac{1}{k+1})$?
27837
Prove by induction: $$\frac1{n+1} + \frac1{n+2} + \cdots +\frac1{3n+1} &gt; 1$$  adding $1/(3m+4)$ as the next $m+1$ value proves pretty fruitless. Can I make some simplifications in the inequality that because the $m$ step is true by the inductive hypothesis, the 1 is already less than all those values?
10408
Prove that for any natural number $n&gt;1$, we have $(2n)!&lt;(n(n+1))^n$    I tried using induction, but I failed in that approach, I rather found it was untrue, but on several case testing, I found it to be true. I am out of ideas here, please help. Thank you.