Prove that $2n-3\leq 2^{n-2}$ , for all $n \geq 5$ by mathematical induction  I have to prove by mathematical induction that: $2n-3\leq 2^{n-2}$ , for all $n \geq 5$
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Prove that $2n-3\leq 2^{n-2}$ , for all $n \geq 5$ by mathematical induction  I have to prove by mathematical induction that: $2n-3\leq 2^{n-2}$ , for all $n \geq 5$  Thank you for the Review.
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I was trying to prove this inequality using induction, but couldn't do.  Question: Suppose $a$ and $b$ are real numbers with $0 &lt; b &lt; a$. Prove that if $n$ is a positive integer, then:  $$a^n-b^n \leq na^{n-1}(a-b)$$
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Prove that for all integers $n$, $n \geq 1$,   $$1 + 3 + 5 + \cdots + (2n - 1) = n\cdot n$$   How would I go about proving this?
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Prove by mathematical  induction that:   $$\forall n \in \mathbb{N}: 3^{n} &gt; n^{3}$$  Step 1: Show that the statement is true for $n = 1$:  $$3^{1} &gt; 1^{3} \Rightarrow 3 &gt; 1$$  Step 2: Show that if the statement is true for $n = p$, it is true for $n = p + 1$  The general idea I had was to start with $(p+1)^{3}$ and during the process substitute in $3^{p}$ for $p^{3}$ as an inequality.  $$(p+1)^{3} = p^{3} + 3p^{2} + 3p + 1 &lt; 3^{p} + 3p^{2} + 3p + 1$$  Now, if it can be shown that:  $$\forall p \in \mathbb{N}: 3p^{2} + 3p + 1 \leq 2 \cdot 3^{p}$$  ...the proof is complete. This is because $3^{p+1} = 3 \cdot 3^{p}$ and one of those three have already been used.   We do this by mathematical induction. First, the base case of $n = 1$:  $$3\cdot 1^{2} + 3 \cdot 1 + 1 \not \leq 2 \cdot 3^{1}$$  ..which turns out to be false.  What are some more productive approaches to this step?
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So I need to prove the inequality :  $$n &lt; 2^n$$  by Induction.  What I have done so far is :     Step $1$: Prove that the statement is true for $n=1$   $$1&lt;2^1$$ (true)      Step $2$:    Prove that, if $p(n)$ is true, then $p(n+1)$ is also true.      Assume that $p(n): n &lt; 2^n$ is true. then :      (Add $1$ on both side to get $n+1$) : $$n+1 &lt; 2^n +1$$      (Multiply the original inequality by $2$ to get $2^{n+1}$ ) :   $$2^{n+1} &lt; 2^n . 2$$   I don't know where to go from here. In fact, I am not sure whether what I did is true or not. I appreciate any insight and help.
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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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I am currently helping a friend of mine with his preperations for his next exam. A big topic of the exam will be induction, thus I told him he should practice this a lot. As at the beginning he had no idea how induction worked, I showed him some typical examples.  Now he showed me an exercise he was having trouble with, which states that one should prove $$n^{n+1} &gt; (n+1)^n$$  for all $n \geq 3$. I have to admit that I too have trouble showing this inequality, as in all of my attempts, my lower bound is too low. Also, I have not yet figured out, how one gets the $(n+2)^{n+1}$, primarely the number $2$ is a problem. I think this might be solved using the binomial theorem, however, I don't think they have already seen the binomial theorem in school.  Is there an easy method to show this inequality by induction not using the binomial theorem? If no: How can one show it using the binomial theorem?  Thanks for answers in advance.
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Show that $$n! \leqslant \left(\frac{n+1}{2}\right)^n \quad \hbox{for all } n \in \mathbb{N}$$  I know that it can be done by induction but I always find line where I do not know what to do next.
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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 $n$.
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I have derived the inequality if $k&gt;1$,  ${n(n-1)⋯(n-k+1)\over k!} ({1\over n})^k&lt;{(n+1)n⋯(n-k+2)\over k!} ({1\over n+1})^k$  But, my problem is how to use this inequality to prove that if $n\geq1$, $(1+{1\over n})^n&lt;(1+{1\over n+1})^{n+1}$
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So far I've proved the following:   $2n \geq \ln(2n) &gt; \ln(n+2)$   for $n \geq 3 $  The proof must be without the use of calculus but I have no idea on how to proceed.
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"Prove that for every $ n $ natural, $24\mid n(n^2-1)(3n+2)$"  Resolution: $$24\mid n(n^2-1)(3n+2)$$if$$3\cdot8\mid n(n^2-1)(3n+2)$$since$$n(n^2-1)(3n+2)=(n-1)n(n+1)(3n+2)\Rightarrow3\mid n(n^{2}-1)(3n+2)$$and$$8\mid n(n^{2}-1)(3n+2)?$$$$$$ Would not, ever succeeded without the help of everyone that this will post tips, ideas, etc., etc.. Thank you.
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I am trying to use mathematical induction to prove  $$(2n)!\ge2^n(n!)^2\quad\text{for }n\in\mathbb{N}$$  I am stuck at the $n=k+1$ point.
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The main question is :  Prove that $n^n&gt;1\times3\times5\times7\times\dots\times(2n-1)$  My approach :  We can write the R.H.S as,  $$\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.
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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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Prove that  $ n &lt; 2^{n}  $ for all natural numbers $n$.  I tried this with induction: Inequality clearly holds when $n=1$. Supposing that when $n=k$, $k&lt;2^{k}$. Considering $k+1 &lt;2^{k}+1$, but where do I go from here?  Any other methods maybe?
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$n\in \Bbb N$ Prove that if $n^2$ is divided by 3, then also n can also be divided by 3.  I started solving this by induction, but I'm not sure that I'm going in the right direction, any suggestions?