Using mathematical induction, prove that $n\le5: 4n&lt;2^n$  base case: $4(5) &lt; 2^5$ $20 &lt; 32$ Correct  I need help with the inductive process
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I did the following base case $n = 5$ $$\begin{align*}     4(5) &amp;\lt 2 ^5\\     20 &amp;\lt 32 \end{align*}$$ So true.  $$\begin{align*}     4n &amp;\lt 2^n\\ n &amp;\lt 2^{n-2}\\ \log_2(n)+2 &amp;\lt n \end{align*}$$ But I don't think this is right. Where do I add in the $(n+1)$.
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 ?
121674
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 have a seemingly simple question to solve by induction. The question says $F_n$ divides $F_{2n}$ in the Fibonacci sequence.  My thoughts on this.   First,since the n-th term of the Fibonacci sequence is given as the sum of the two previous terms, simple induction won't suffice. So I should try composite induction.   That is, supposing that $F_k$ divides $F_{2k}$ for all $k&lt;n$, I must prove $F_{n}$ divides $F_{2n}$. Is this approach valid?  Doesn't seem to work. If I write $F_{2n}=F_{2n-1}+F_{2n-2}$ then by the inductive hypothesis, $F_{n-1}$ divides $F_{2n-2}$ but this does not help me assert that $F_n$ divides $F_{2n}$. Am I doing this wrong?
123027
I am having trouble with my homework problem, it says:      Suppose that $n$ is a positive integer and that $x &gt; 0$. Show that   $$\left(1+\frac{x}{n}\right)^n &lt; e^x.$$    I have proved the base case, but I am unsure how to proceed with the induction step. The question continues:      and, if $x &lt; n$, then   $$e^x &lt; \left(1-\frac{x}{n}\right)^{-n}$$   Also, by choosing a suitable $n$, deduce that $2.5 &lt; e &lt; 2.99$.   I would like a hint to help me along if anyone can help. Thanks!
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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?
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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.
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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.
72107
For $n \geq 2$ prove that $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$  We need to use induction.   The Principle of Mathematical Induction, Theorem 4.2.1, states that $n_0 \in \mathbb{Z}$. For each integer $n \geq n_o$, let $P(n)$ be a statement about $n$. Suppose that the following two statements are true:   $P(n_o)$ $( \forall n \geq n_o)[P(n) \rightarrow P(n+1)]$   Then, for all integers $n \geq n_o$, that statement $P(n)$ is true.  Suppose,  $(1- \frac{1}{4})(1- \frac{1}{9})(1- \frac{1}{16})...(1- \frac{1}{n^2}) = \frac{n+1}{2n}$, the equation, is true for some natural number $k$.  $\frac{1}{2^2})*...*(1- \frac{1}{k^2}) = \frac{k+1}{2k}$  The induction hypothesis $P(k) \rightarrow P(k+1)$ must hold for any natural number $k, P(k),$ from the basis and inductive step.  $(\frac{1}{2^2})*...*(1- \frac{1}{k^2})*(1- \frac{1}{(k+1)^2}) = \frac{k+1}{2k}(1- \frac{1}{(k+1)^2})$  $\leftrightarrow$ $\frac{k+1}{2k}(\frac{(k+1)^2}{(k+1)^2}- \frac{1}{(k+1)^2})$  $\leftrightarrow$ $(\frac{(k+1)}{2k})( \frac{(k+1)^2-1}{(k+1)^2})$  $\leftrightarrow$ $( \frac{(k^2+2k+1-1}{(2k)(k+1)})$  $\leftrightarrow$ $( \frac{(k^2+2k}{(2k)(k+1)})$  $\leftrightarrow$ $( \frac{k+2}{(2)(k+1)})$  We need to demonstrate that   $(\frac{1}{2^2})*...*(1- \frac{1}{n^2}) = \frac{n+1}{2n}$  holds true for $n \geq 2$  The question is do I just substitute a number that is greater than two after I did the induction?
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?
122259
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.
94204
As stated in the tag, I'm trying to prove by induction the claim $F_{2n} = F^2_{n+1} - F^2_{n-1}$, where $F_{n}$ is the $n^{th}$ Fibonacci number. I've spent hours on the inductive step without substantial progress, and am hoping someone can provide a path to the desired result.    To facilitate the proof, we're given $F_{2n-1} = F^2_{n} + F^2_{n-1}$  (Also, in various past attempts not reflected here I tried to use the following substitutions whenever it seemed prudent: $F_{n} = F_{n-1} + F_{n-2} = F_{n+1} - F_{n-1}$; and $(x^{2} - y^{2}) = (x - y)(x + y)$).  For the inductive step, let $n\geq 1$ and assume $F_{2n} = F^2_{n+1} - F^2_{n-1}$. For $n+1$ we want to show $F_{2(n+1)} = F^2_{n+2} - F^2_{n}$. So  \begin{align}   F_{2(n+1)} &amp;= F_{2n + 2} \\              &amp;= F_{2n+1} + F_{2n} \\              &amp;= (F_{2n} + F_{2n-1}) + F_{2n} \\              &amp;= 2 F_{2n} + F_{2n - 1} \\              &amp;= 2 (F^2_{n+1} - F^2_{n-1})                   + F^2_{n} + F^2_{n-1}                       &amp;&amp;\text{by the inductive hypothesis} \tag{1} \\              &amp;=...F^2_{n+2} - F^2_{n}                      \end{align} I have two challenges.  The first is squeezing $-2 F^2_{n}$ out of (1).  The only reasonable way I can see to do that is to expand the $-F^2_{n-1}$ term in  $2 F^2_{n+1} - F^2_{n-1}$ to $-(F_{n} - F_{n-2})^2$. If in instead I combine it with $F^2_{n-1}$ then it seems impossible to generate the $-2 F^2_{n}$ term needed to offset $F^2_{n}$ and produce $-F^2_{n}$.   The second challenge is to find $(F^2_{n+2}$ in (1), and the only way I can see to do that is to expand the $F^2_{n+1}$ term in $2(F^2_{n+1} - F^2_{n-1})$to $(F_{n+2} - F_{n})^2$. Doing that not only generates $2 F^2_{n}$ which totally screws things up, but it also generates a bunch of intermediate terms I can't get rid of, and it's driving me crazy.
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I have difficulty in neatly writing down a proof for the following from Terence Tao's Analysis I, where $m{+\!+}$ is the :     Let $n$ be a natural number, and let $P(m)$ be a property pertaining to the natural numbers such that whenever $P(m{+\!+})$ is true, then $P(m)$ is true. Suppose that $P(n)$ is also true. Prove that $P(m)$ is true for all natural numbers $m\leq n$; this is know as the principle of backwards induction. (Hint: apply induction to the variable $n$.)   First of all, I am unsure about what the base case should look like. For the induction step, I understand that if we suppose inductively that $P(n)$ is true, that then for a natural number $a$ s.t. $a{+\!+}=n$ it holds that $P(a)$ is true, and then for a natural number $b$ s.t. $b{+\!+}=a$ it holds that $P(b)$ is true etc. Hence for all natural numbers $m\leq n$, $P(m)$ is true.  Could anyone please tell me what the base case should look like, and whether there is a neater way of writing down the induction step?
96833
How does one prove by induction that  $n! &gt; n^2$  for $n \geq 4$
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I came across the following in the book Handbook of Mathematical Induction: $$ 19\mid (2^{2^n}+3^{2^n}+5^{2^n}),\quad n\in\mathbb{Z^+}\tag{1} $$ Apparently, this problem is not so bad if you think about it in a somewhat "outside of the box" sort of way, but it is supposed to be quite challenging otherwise.   Question: How might one prove $(1)$ using induction or otherwise? The "outside of the box" answer has been provided but with details oranged out so as not to spoil it for anyone who may come up with that solution individually.
60331
I attempted a proof of mathematical induction using an arbitrary base case, but was unsuccessful (and hence this question). Below is what I was trying to do and along with my thinking; if anyone can point me in the right direction I'd appreciate it.  The induction I am using: Let $P$ be a property about the natural numbers $\mathbb{N}$ with $0\in\mathbb{N}$, and let $P(n)$ denote the statement that the property $P$ holds for $n\in\mathbb{N}$. Suppose $P(0)$. Furthermore suppose that for each natural number $k$, $P(k)$ implies $P(k+1)$. Then $\forall nP(n)$.  What I am trying to prove: Let $P$ be a property about the natural numbers $\mathbb{N}$ with $0\in\mathbb{N}$, and let $P(n)$ denote the statement that the property $P$ holds for $n\in\mathbb{N}$. Suppose for $n_0\in\mathbb{N}$, $P(n_0)$. Furthermore suppose that for each natural number $k\geq n_0$, $P(k)$ implies $P(k+1)$. Then $\forall n\geq n_0 P(n)$.  My attempted proof: define $Q(n)$ to be $n\geq n_0 \to P(n)$. Then we wish to prove $\forall nQ(n)$. We induct on $n$.  Base case (for ordinary induction): $Q(0)$ is $0\geq n_0\to P(0)$. Since $n_0\in\mathbb{N}$, $0\geq n_0$ implies that $n_0=0$. Since $P(n_0)$, $P(0)$, which proves the base case.  Inductive step: we want to show $\forall n(Q(n)\to Q(n+1))$. To do this, we assume $k\geq n_0 \to P(k)$ and try to show $k+1\geq n_0 \to P(k+1)$.  Since $k\geq n_0 \to P(k)$, we first prove the case where $k\geq n_0$ and $P(k)$. By the hypothesis of the proof, we see that $P(k+1)$, which proves the case for $k+1$.  This is where I am having trouble: For $k&lt;n_0$, I am unable to show the implication for $k+1$.  So my questions would be: (1) is the overall approach for the proof correct? (2) If so, how might I go on to prove the case when $k&lt;n_0$?  Thanks in advance. (This is not homework, by the way.)
23125
I aim to show that $$(n+1)^n&lt;n^{(n+1)}$$ for all $n \geq 3$. I tried induction, but it didn't work. What should I do?