I would like to do a proof by induction. I know there's a few duplicate posts about this question but I dont quite understand the notation they have given while showing an induction proof. If someone could post an induction proof without any complex notation for the product of n consecutive integers, I would really appreciate it.    - this answer, how does $(𝑚+1)(𝑚+2)...(𝑚+𝑘+1) = (k+1) (m+1)...(m+k) + m^{k+1}$ Can anyone expand this notation to show me whats happening?
28821
How can we prove that the product of $n$ consecutive integers is divisible by $n$ factorial?     Note: In this  and the comments here the OP has clarified that he seeks a proof that "does not use the properties of binomial coefficients". Please post answers in said newer thread so that this incorrectly-posed question may be closed as a duplicate.
96688
My question concerns the inductive step of the proof below. I've typed out the work I've done so far. If anyone has any suggestions on how to proceed thanks in advance.  If $n\in\mathbb{N}$, then $(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^n})\ge\frac{1}{4}+\frac{1}{2^{n+1}}$.  $Proof.$ We use mathematical induction.  Base Case. For $n=1$ we see that $1-\frac{1}{2}\ge\frac{1}{4}+\frac{1}{4}$ since $\frac{1}{2}\ge\frac{1}{2}$.  Inductive Step. Let $n&gt;1$. Assume our statement is true for $n$. Now observe that  $\begin{align}(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n+1}})&amp;\ge\\ \left((1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n}})\right)(1-\frac{1}{2^{n+1}})&amp;\ge(\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$  My question is: How do we get from $\begin{align} (\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$ to $\frac{1}{4}+\frac{1}{2^{(n+1)+1}}$?  Most of what I've attempted has led to tedious calculations that do not simplify to the desired result. I've also tried making the argument that $(1-\frac{1}{2^{n+1}})$ must be $\ge$ to $\frac{1}{2}$ but then I'm stuck with $\frac{1}{8}+\frac{1}{2^{(n+1)+1}}$... Is my approach incorrect here?
43934
How do I show via inductive proof (and without the use of the binomial theorem) that $$ {n\choose k} \in {\mathbb{N}} $$ $$\text{ for  all }\: n,k \in \mathbb{N}_0\: \text{ with }\: n≥k .$$ For my induction proof, I would show that$$  {n\choose k-1}+{n\choose k} = {n+1 \choose k}   $$ is true. However, how do I start the induction correctly with the base cases and the hypothesis? As far as I know, I cannot simply assume $$ {n\choose k-1}+{n\choose k}, $$  so how do I come to this derivation?
115600
Use induction to prove that that $8^{n} | (4n)!$ for all positive integers $n$  So far I have: Base case (n = 1) = $8^{1} | (4(1))!$  = $8 | 24$ which is true.  Induction Step:  $8^{n + 1} | (4(n + 1))!$  $8^{n + 1} | (4n + 4)!$   A bit confused as to how to close this proof out, also wanted to make sure my current progress is correct as well. Any help is appreciated.
14779
How would one prove that  $$(n+1)^{n-1}&lt;n^n \ \forall n&gt;1$$  I have tried several methods such as induction.
58390
The following problem:  The Fibonacci number is denoted by  $f_{n}$, show that the following holds when $n$ is a positive integer:  $f_{0}f_{1}+f_{1}f_{2}+...+f_{2n-1}f_{2n} = f^{2}_{2n}$  My idea was to do an induction proof, which holds for $n$ = 1  $f_{0}f_{1} + f_{1}f_{2} = f^{2}_{2}$  So we assume it holds for $n$ and show for $n+1$  $f_{0}f_{1}+f_{1}f_{2}+...+f_{2n-1}f_{2n} + f_{2n}f_{2n+1} = f^{2}_{2n} + f_{2n}f_{2n+1}$  I fail to get  $f^{2}_{2n+1}$ out of the term  $f^{2}_{2n} + f_{2n}f_{2n+1}$  Is there a difference between "show that" and "prove that". So might be induction the wrong tool here?  Many thanks for your hints in advance :-)
62381
Can you please help me and tell, how should I move on?  Can this be proved by induction?     Every natural number $n\geq 8$ can be represented as $n=3k + 5\ell$.   Thank you in advance
5493
I need to prove the following, but I am not able to do it. This is not homework, nor something related to research, but rather something that came up in preparation for an exam.     If $n = 1 + m$, where $m$ is the product of four consecutive positive   integers, prove that $n$ is a perfect square.   Now since $m = p(p+1)(p+2)(p+3)$;     $p = 0, n = 1$ - Perfect Square      $p = 1, n = 25$ - Perfect Square      $p = 2, n = 121$ - Perfect Square   Is there any way to prove the above without induction? My approach was to expand $m = p(p+1)(p+2)(p+3)$ into a 4th degree equation, and then try proving that $n = m + 1$ is a perfect square, but I wasn't able to do it. Any idea if it is possible?
123714
Question: Let $m,n\in\mathbb{N}$, prove that if $n|m$, $F_n|F_m$.  I've tried to use induction, but I don't really know where to start since there's $2$ numbers:  $n$ and $m\ \dots$ I did induction before with just $1$ number, like proving $1+2+\cdots+n=\frac{n(n+1)}{2}$, that's only dealing with one number $n$, but with this I have no clue how to do it.  Can anyone give me a little hint on how to start, I don't want you to do whole problem for me but can I have some hint so I know where to start?  Thanks for the help!
9434
Using induction, how can I show the following identity about the fibonacci numbers? I'm having trouble with simplification when doing the induction step.  Identity: $$f_n^2 + f_{n+1}^2 = f_{2n+1}$$  I get to:  $$f_{n+1}^2 + f_{n+2}^2$$  Should I replace $f_{n+2}$ using the recursion? When I do that, I end up with the product of terms, and that just doesn't seem right. Any guidance on how to get manipulate during the induction step?  Thanks!
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.
115709
Proving by induction. We'd like to show that $2 + 4 + 6 + \cdots+ 2n = n(n + 1)$.  A nice way to do this is by induction. Let $S(n)$ be the statement above. An inductive proof would have the following steps:  Show that $S(1)$ is true.  Show that if $S(1),\ldots,S(k)$ are true, then so is $S(k + 1)$.    This question is really starting to bug me, am I doing something wrong or is the equation wrong for that series? It only seems to work for $S(1)$ but after that it does not give the correct series. The $n(n+1)$ series is $2, 6, 12,\ldots.$ I'm really confused can someone please nudge me in the right direction (I know how proofs by induction works) I'm just having problems with this one in particular.
67724
Provided the following sequence of integers $t_1, t_2, t_3$,... is defined as:    $t_1 =4, t_2 =8$ and $t_n= $ $ 6t_n$$_-$$_1$ - $4t_n$$_-$$_2$ for all integers $n \geq 3$   How do we prove that $2^n$ divides $t_n$ for all integers $n \geq 1$ by strong induction?  So far, I am having problems with the induction step as I am still trying to grasp the concept of strong induction, but this is what I came up with so far:  Base step:  $t_3 = 32$, which is divisible by 8; and $t_4 = 160$, which is divisible by 16.  $\therefore$ Statement is true when $n=3$ and $n=4$.  Induction step:  Assuming $k=n$ and the statement is true for $3,4,...k$. Prove that $2^k$ divides $t_k$ for $k+1$:  $t_3 = ((2^3 - 2)(2^3) - (2^2)(2^2)) = 2^6 - 2^4 - 2^4 $; dividing this by $2^3$ gives $2^3 - 2^2$ which would be an integer result.  $t_4 = ((2^3 - 2)(2^5) - (2^2)(2^3)) = 2^8 - 2^6 - 2^5 $; dividing this by $2^4$ gives $2^4 - 2^2 - 2$ which would also be an integer result.  $\therefore$ By the logic above, the statement also holds for $k-1$, $k$ as well as $k+1$, thus as required to prove by strong induction.  I believe this is a fundamentally flawed way of trying to prove by strong induction, but this is the only way I can possibly think of. Can someone show a correct method of solving this?
44709
I'm trying to prove using induction that $9$ divides $n^3 + (n+1)^3 + (n+2)^3$ whenever $n$ is a non-negative integer. So far, I have:   Base case: $P(1) = (1) + (8) + (27) = 36, 36$ can be divided by $9$ so the base case is valid Inductive step: let $P(n)$ be the statement $9$ divides $n^3 + (n+1)^3 + (n+2)^3$. Assume $P(k)$ is true, so $9$ divides $k^3 + (k+1)^3 + (k+2)^3$.   And this is where I'm stuck. I'm not sure how to demonstrate that $9$ divides $P(n)$ when $n = k+1$. If someone could step me in the right direction that would be awesome. Thanks!
86171
I'm supposed to prove that, for every integer $n &gt; 0,$ it is true that $(1 + 2 + ... + n)$ divides $3(1^2 + 2^2 + ... + n^2)$.  Should I use induction?    This was given as an exercise in a chapter about divisibility in my algebra textbook.
54219
I'm not too sure exactly how to approach this question. Would anyone be able to give me any helpful advice or some sort of direction? I have a little problem with induction.   Prove by induction that:  $$1 + 3 + 5 + 7 +...+ (2n+1) = (n+1)^2 \quad\forall n \in \mathbb N $$
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.