Am sorry if i might seem a little noob, but my lecturer taught us to prove a few proof of functions by induction and even gave a few examples e.g $1 + 3 + 5 + 7 + ... + 2n-1 = n^2$  But seriously he did'nt event tell us what proving a function by induction is. Please if someone would provide a brief and well understandable intro to proving functions by induction is. I'd appreciate! And to use as an example, is that equation there. Thanks in advance
121306
What does  "Prove by induction" mean? Would you mind giving me an example?
67186
I have to do proof of $(a+b)^n$ and $(a+b)^{n+1}$ with mathematical induction.  I finished the first one $(a+b)^n = a^n + na^{n-1}b+\dots+b^n$.  I however have trouble with the second one, I don't know what to start exactly. Any help/hints?
3295
Proofs that proceed by induction are almost always unsatisfying to me. They do not seem to deepen understanding, I would describe something that is true by induction as being "true by a technicality". Of course, the axiom of induction is required to pin down the Natural numbers and certainly seem to be indispensable in one form or another.   However, I am still interested in the following: Are there any "natural" theorems in mathematics that seem unlikely to fall to any method other than induction for whatever reason? I would not include examples that proceed by breaking down a structure into smaller components that are more easily handled - somehow these proofs satisfy whatever criteria for beauty I have in my mind.   An example of a theorem that does have an inductive proof and a more "superior" proof is Fermat's little theorem. It is perfectly possible to prove it by induction but the proof through group theory seems better - perhaps because it is more easily generalizable. I would like examples where it seems like the "neat" proof is unlikely to exist.  This is probably very philosophical and I do not really have a concrete question but I am sure I am not alone in feeling this way.
12895
So what I am trying to prove is for any real number x and natural number n, prove $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$   I think that to prove this I should use induction, however I am a bit stuck with how to implement my induction hypothesis. My base case is when $n=2$ we have on the left side of the equation $x^2-1$ and on the right side: $(x-1)(x+1)$ which when distributed is $x^2-1$. So my base case holds.   Now I assume that $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ for some $n$. However, this is where I am stuck. Am I trying to show $x^{n+1}-1=(x-1)(x^n + x^{n-1}+x^{n-2}+...+x+1)$? I am still a novice when it comes to these induction proofs. Thanks
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 ?
24893
Prove by induction that $\forall n \in\mathbb N$, $3\mid4^{n}-1$.  1) For $n = 1$ the statement is obviously true.  2) Now what about $n+1$? I was thinking of writing $4^{n}-1$ as $2^{2n}-1$ and then $4^{n+1}-1 = 2^{2n+2}-1$ but that got me nowhere.
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.
25374
I'm trying to prove by induction that for all $a,b \in \mathbb{Z}$ and $n \in \mathbb{N}$, that $(a-b) \mid (a^n-b^n)$. The base case was trivial, so I started by assuming that $(a-b) \mid (a^n-b^n)$. But I found that: \begin{align*} (a-b)(a^{n-1}+a^{n-2}b + a^{n-3}b^2+...+b^{n-1}) &amp;= a^n-b^n. \end{align*}  Doesn't this imply that $(a-b) \mid (a^n-b^n)$ as $a^{n-1}+a^{n-2}b+\dots+b^{n-1}$ is clearly an integer? This obviously isn't a proof by induction, but is there anything wrong with taking this approach to prove this result, other than the fact that it isn't what is being asked?   Aside, I'm having trouble getting started with the proof by induction. I've tried making the above equivalence useful during the induction step, but it doesn't seem to be helpful, so any hints would be greatly appreciated!
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.
18754
I just realized something interesting. At schools and universities you get taught mathematical induction. Usually you jump right into using it to prove something like   $$1+2+3+\cdots+n = \frac{n(n+1)}{2}$$  However.  I just realized that at no point is mathematical induction proved itself? What's the mathematical induction's proof? Is mathematical induction proved using mathematical induction itself? (That would be mind blowing.)
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 :-)
99260
So I see that the process for proof by induction is the following (using the following statement for sake of example:  $P(n)$ is the formula for the sum of natural numbers $\leq n$:  $0 + 1 + \cdots +n = (n(n+1))/2$ )   Show that the base case is trivially true: $P(0)$: let $n = 0$. Then $0 = (0(0+1))/2$, which is true. Show that if $P(k)$ holds then $P(k+1)$ also holds. Assume $P(k)$ holds for some unspecified value of $k$. It must be then shown that $P(k+1)$ holds.   This is the part I don't get: 'Assume $P(k)$ holds'  We are just assuming something is true and then 'proving' that something else is true based on that assumption. But we never proved the assumption itself is true so it doesnt seem to hold up to me. Oviously proof by induction works so am I viewing the process incorrectly?
67175
I am having trouble to understand the way of using a induction for the following example:     Let $\Sigma \overset{\Delta} = \{a, b\}$ and $S_1 \overset{\Delta} = \{a^n \mid n \in \Bbb N\}$. Prove using induction that $$\forall w \in \Sigma^* . w \in S_1 \to \big( w = \lambda \lor (\forall \ 1 \le i \le |w| . (w)_i = a) \big) .$$   How to do that? Any help is upvoted.
12773
Use mathematical induction to prove this. Here is my answer but I stuck at certain point.   Base Case:           n=1  $$(1+x)^1 \ge 1+x $$ True ,   Induction Case:           n=k assume    $$(1+x)^k \ge 1+kx $$           n=k+1           $$ (1+x)^k+1 \ge 1+(k+1)x $$           $$(1+x)^k *(1+x) \ge 1+ kx+ x $$        Stuck!!!
103794
I apologize in advance for asking such elementary question but I would like to show that  $$(i+k)! &gt; i!k!$$ I believe induction can prove it but is there an algebraic proof on proving about it?
10141
Prove: $\binom{n}{0}F_0+\binom{n}{1}F_1+\binom{n}{2}F_2+\cdots+\binom{n}{n}F_n=F_{2n}$;  I was stuck with this question for a while... Help me please!!! Thanks!!!
62511
When doing a proof by mathematical induction, I was wondering if there is any logical reason why the assumption (n=k) and induction (n=k+1) steps couldn't be done first, then do the basis case (n=1) afterwards, rather than the traditional way of doing this basis case first? I am teaching this to my students and I think it would make more sense to them to first show that P(k+1) is true if P(k) is true, and then show that P(1) is true, then P(2) must be true, P(3) is true, etc. and P(n) is true for all n. Anyway, I can't see any flaws in the logic? Please advise.