Prove using mathematical induction that $n&lt;3^n$ for all positive integers $n$ I came until here, 3^k+1&lt; 3^(k+1) I wanna know if this is already the final answer?
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As title says :     Prove using mathematical induction that $n &lt; 3^n$ for all positive integers $n$.   I came till here  $(k+1)&lt;3^k + 1$  From here I don't have any idea how to show $(k+1)&lt;3^{(k+1)}$ . Please help.
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How can I prove that $n!&lt;n^n$ for every $n&gt;1$ using the induction principle?  If I put $n=2$ I get $2&lt;4$ so I know that $n!&lt;n^n$ is true.  Now I don't know how to prove that $(n+1)!&lt;(n+1)^{n+1}$. I can write $(n+1)!=n!(n+1)$ but then I don't know how to go on.
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One of my Fellows asked me whether total induction is applicable to real numbers, too ( or at least all real numbers ≥ 0) . We only used that for natural numbers so far.  Of course you have to change some things in the inductive step, when you want to use it on real numbers.  I guess that using induction on real numbers isn't really possible, since $[r,r+\epsilon]$ with $\epsilon > 0$, $r \in \mathbb R$ is never empty.  Can you either give a good reason, why it isn't possible or provide an example where it is used?
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I am trying to prove that   $$e^{k+1} ≥ 3 + 3k + k^2$$ with, $$k&gt;2$$  WhatI have done so far:  What we are trying to prove is that $$e^n≥1+n+n^2$$ is a true statement. Since $n=3$ holds, this is our base case. Then, the inductive hypothesis is: $$e^k≥1+k+k^2$$ Hence, I need to show $$e^{k+1}≥1+(k+1)+(k+1)^2=3+3k+3k^2$$ is a true statement.
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I'm trying to prove that for $a,b\in \mathbb{R}^+$, if $a&lt;b$, then $a^{1/n} &lt; b^{1/n}$. I'm naturally proceeding by induction on $n$, with it being trivial for $1$. I've proven it separately before with $n=2$ as well, and I'm just gong to assume it true here for convenience. My problem is in the induction step, having it assumed to be true for some fixed $n$. I need to show $a^{1/(n+1)} &lt; b^{1/(n+1)} $. I'm having a complete brain fart. Can anyone help?
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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
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I need to prove that $$ 2^n &gt; n^3\quad \forall n\in \mathbb N, \;n&gt;9.$$  Now that is actually very easy if we prove it for real numbers using calculus. But I need a proof that uses mathematical induction.   I tried the problem for a long time, but got stuck at one step - I have to prove that:     $$ k^3 &gt; 3k^2 + 3k + 1 $$   Hints???
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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!
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I've been dealing with this problem for almost 2 hours now, with hardly any progress. I'm to prove the following inequality using induction   $$\sqrt[n+1]{n+1}&lt;\sqrt[n]{n}$$ where $n≥3$, $n∈\mathbb{N}$. How can this be done?
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I am looking for an example where you have $P(n)$ implying $P(n+1)$. However there is no base case. For which there is therefore no solution at all for the induction problem even though the inductive step itself works.
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Use Mathematical induction to prove that for all integers, $n$ is greater than or equal to $1$. I am confused on what to do after I do the the basis step that is using $n$ as $1$.   $$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} +...+\frac{1}{n(n+1)}=1-\frac{1}{n+1}$$
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Prove that for any natural number n the following equality holds:   $$ (1+2+ \ldots + n)^2 = 1^3 + 2^3 + \ldots + n^3 $$  I think it has something to do with induction?
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I'm trying to freshen up for school in another month, and I'm struggling with the simplest of proofs!     Problem:      For any natural number $n , n^3 + 2n$ is divisible by $3.$   This makes sense      Proof:      Basis Step: If $n = 0,$ then $n^3 + 2n = 0^3 +$   $2 \times 0 = 0.$ So it is divisible by $3.$      Induction: Assume that for an arbitrary natural number $n$,   $n^3+ 2n$ is divisible by $3.$       Induction Hypothesis: To prove this for $n+1,$ first try to express $( n + 1 )^3 + 2( n + 1 )$ in terms of $n^3 + 2n$ and use   the induction hypothesis. Got it      $$( n + 1 )^3+ 2( n + 1 ) = (  n^3 + 3n^2+ 3n + 1 ) + ( 2n + 2 ) \{\text{Just some simplifying}\}$$      $$ = ( n^3 + 2n ) + ( 3n^2+ 3n + 3 ) \{\text{simplifying  and regrouping}\}$$   $$ = ( n^3 + 2n ) + 3( n^2 + n + 1 ) \{\text{factored out the 3}\}$$      which is divisible by $3$, because $(n^3 + 2n )$ is divisible by $3$   by the induction hypothesis. What?   Can someone explain that last part? I don't see how you can claim $(n^3+ 2n ) + 3( n^2 + n + 1 )$ is divisible by $3.$
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These are just the examples from my textbook, but I don't think it did not explain well.  One of the problem was to prove the inequality   $$n&lt;2^n$$  for all integers $n$.  I understand we assume if $P(n)$ is true, $P(k+1)$ is true.  Thus $k+1&lt; 2^{k+1}$  But in the text book, they added one more line that I don't understand:  $$k+1&lt;2^k+1 \le 2^k+2^k=2\cdot2^k=2^{k+1}$$  I understand  $$k+1&lt;2^{k+1}$$  But how do I know $2^k+1$ is larger than $k+1$, and smaller than $2^{k+1}$?  I have one another example just in case,  proving $2^n&lt;n!$ using induction.  $$2^{k+1}=2\cdot 2^k&lt;2\cdot k!&lt;(k+1)k!=(k+1)!$$  I understand we should start with $2^{k+1}&lt;(k+1)!$, but  Where did $2\cdot k!$ come from? Textbook says it came from inductive hypothesis, but I still don't get it.  Please advise.
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Prove that $n&lt;3^n$ where $n \in \mathbb N$,  when $ n=1 $, I have proved it's true. And assumed when $n = p$ , $p&lt;3^p$ is true. Can any body help me in showing that it is true for $n =p+1$
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How would you go about proving $n! &lt; n^n$ using a mathematical induction proof?  I understand how to solve inductive proofs with = but I'm getting a bit lost in this example.   Any help is much appreciated.
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I am not sure whether this is right. Can anyone verify, whether this proof is valid? Thanks!        Define a sequence $\{a_n\}_{n\ge0}$ as follows:   $$a_0=1,\qquad,a_1=3,\qquad,a_2=9,\qquad,a_n=a_{n-1}+a_{n-2}+a_{n-3}\text{ for }n\ge3.$$   Prove that for any positive integer $n$, $a_n\le3^n$.        Proof. Let $P(n)$: $a_n\le 3^n$, $n$ is a non-negative integer.      (i) Base case:         Consider when $n=0$. LHS$=a_0=1$, RHS=$3^0=1$ $\therefore$ LHS$\le$RHS, then $P(0)$ holds.   Consider when $n=1$. LHS$=a_1=3$, RHS=$3^1=3$ $\therefore$ LHS$\le$RHS, then $P(1)$ holds.   Consider when $n=1$. LHS$=a_2=9$, RHS=$3^2=9$ $\therefore$ LHS$\le$RHS, then $P(2)$ holds.         (ii) Inductive case:   Assume $P(i)$ is true for $0 \le i \le k$, $k\ge2$.      (iii) Inductive conclusion: Consider $n=k+1$.   $$ \begin{align*} \mathrm{LHS}=a_{k+1}&amp;=a_k+a_{k-1}+a_{k-2} \qquad\text{(by definition, since $k\ge 2$)}\\                     &amp;\le 3^k+3^{k-1}+3^{k-2} \qquad\text{(by Induction Hypothesis)}\\                     &amp;=3^k(3^{-1}+3^{-2}+3^{-3})\\                     &amp;=3^k\left(\frac13+\frac19+\frac1{27}\right)\\                     &amp;=\frac{13}{27} 3^k\\                     &amp;\le 3^{k+1} = \mathrm{RHS} \end{align*} $$   Therefore, by the Principle of Strong Induction, $P(n)$ is true for all non-negative positive integers $n$.