Using induction, prove $|\sin(nx)| \le n|\sin x|$ for $n\in \mathbb{N}$.    How I can show that with induction? Normally , When I make a induction I know for $n$, and I prove for $n+1$. But I can't write $n+1$ in this question. If I write $n+1$, formula change like this: $|\sin((n+1)x)| \le (n+1)|\sin x|$ for $n+1$, This new formula doesn't include $n$.   What should I do in such cases?
123466
I have to show that for every $x \in \mathbb{R}$ and every $n \in \mathbb{N}$ $$|\sin(nx)| \le n|\sin(x)|  $$ In the previous exercise, I have showed that $|\sin(x)|≤|x|$ with the use of the mean value theorem. I think that I cannot use this approach this time. I also tried to write $\sin(nx)$ as a series expansion but that doesn't work either.     Does anyone know how I can solve this?
88669
Let $n$ and $r$ be positive integers with $n \ge r$. Prove that   Still a beginner here. Need to learn formatting.  I am guessing by induction? Not sure what or how to go forward with this.  Need help with the proof.
81069
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 $1$.   $$\frac{1}{1 \cdot 2} + \frac{1}{2 \cdot 3} +...+\frac{1}{n(n+1)}=1-\frac{1}{n+1}$$
45547
Prove by induction    $$ \sin x + \sin 2x + ... + \sin nx = \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} $$   What I have for now: $$  \frac {\sin (\frac {n + 1} {2} x)} {\sin \frac{x}{2}} \sin \frac{nx}{2} + \sin(n + 1)x =  \frac {\sin (\frac {n + 2} {2} x)} {\sin \frac{x}{2}} \sin \frac{(n + 1)x}{2}$$ Letting $y = \frac {(n + 1)x} {2} $   $$\frac {\sin y} {\sin \frac{x}{2}} \sin (y - \frac{x}{2}) + \sin2y =  \frac {\sin (y + \frac {x} {2} )} {\sin \frac{x}{2}} \sin y $$  Then I used $\sin (\alpha + \beta)$ and $\sin (\alpha - \beta)$ formulas, bit they didn't help.
73759
Prove that $\sqrt{n} &gt; \ln n$ for all $n \in \mathbb{N}$.  I need to use this fact for one of the proofs that I am working on. However, I am having trouble proving this. I tried induction but don't really know how to do it. Can someone help me with this?
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.
110441
I know how to prove this by induction but the text I'm following shows another way to prove it and I guess this way is used again in the future. I'm confused by it.  So the expression for first n numbers is: $$\frac{n(n+1)}{2}$$  And this second proof starts out like this. It says since:  $$(n+1)^2-n^2=2n+1$$  Absolutely no idea where this expression came from, doesn't explain where it came from either.  Then it proceeds to say: \begin{align} 2^2-1^2&amp;=2*1+1 \\ 3^2-2^2&amp;=2*2+1\\ &amp;\dots\\ n^2-(n-1)^2&amp;=2(n-1)+1\\ (n+1)^2-n^2&amp;=2n+1 \end{align} At this point I'm completely lost. But it continues to say "adding and noting the cancellations on the left, we get" \begin{align} (n+1)^2-1&amp;=2(1+2+...+n)+n \\ n^2+n&amp;=2(1+2+...+n) \\ (n(n+1))/2&amp;=1+2+...+n \end{align}  Which proves it but I have no clue what has happened. I am entirely new to these math proofs. Im completely lost. I was great at high school math and calculus but now I haven't got the slightest clue of what's going on. Thanks
68551
Use mathematical induction to show that:  $1^3 + 3^3 + 5^3 + ... + (2n + 1)^3 = (n+1)^2(2n^2+4n+1)$ whenever $n$ is a positive integer.  I am able to solve this problem, but the thing that confuses me on this is the Basis Step.  P(1): $(2(1) + 1)^3$ $\neq$ $4(7)$, does this mean that this problem can not be proved using mathematical induction, or am I missing something? Thanks.
60101
Let $n \in \mathbb N^+$. Show that if $x_1, x_2, ... , x_n$ are $n$ real numbers such that $-1 \le x_i \le 0$ for each $1 \le i \le n$, then $$(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + x_1 + x_2 + ... + x_n$$  I am using a proof by induction, but I am unable to complete it. I would appreciate any tips on how I could finish the proof.   Incomplete Proof Let $P(n)$ be the proposition that $(1 + x_1)(1 + x_2)...(1 + x_n) \ge 1 + \sum_{i = 1} ^ n x_i$. When $n = 1$, LHS = RHS = $1 + x_i$. Clearly $1 + x_i \ge 1 + x_i$ and so $P(1)$ is true. Assume that $P(k)$ is true, i.e. $(1 + x_1)(1 + x_2)...(1 + x_k) \ge 1 + \sum_{i=1}^k x_i$. Case 1 Suppose $x_{k + 1}  = -1$. Then $1 + x_{k + 1} = 0$, hence LHS = $(1 + x_1)(1 + x_2)...(1 + x_k)(1 + x_{k+1}) = 0$. RHS = $1 + x_1 + x_2 + ... + x_k - 1 = x_1 + x_2 + ... + x_k$ which is always nonpositive since each $x_i$ is nonpositive. Hence RHS $\le 0$ and so LHS $\ge$ RHS. Case 2 Suppose $x_{k+1} = 0$. Then $1 + x_{k+1} = 1$, hence multiplying it to LHS does not change anything. Also, adding $x_{k+1}$ to RHS does not change anything. Therefore, LHS $\ge$ RHS still holds. Case 3 Suppose $-1 &lt; x_{k+1} &lt; 0$. Then $0 &lt; 1 + x_{k+1} &lt; 1$. Adding $x_{k+1}$ to RHS makes the RHS smaller since $x_{k+1}$ is negative. If LHS $= 0$, then multiplying $1 + x_{k+1}$ to LHS does not change anything, and so LHS $\ge$ RHS holds. If LHS > $0$, then multiplying $1 + x_{k+1}$ to LHS makes the LHS smaller since $0 &lt; 1 + x_{k+1} &lt; 1$. In this case LHS $\ge$ RHS does not necessarily hold? ...
62722
Use induction to prove that, for all $n \in \mathbb{Z}^+$, $2\mid (n^2 − n)$.   That is, I am supposed to use induction to prove that $(n^2 − n)$ can be divided by $2$ when $n$ is a positive integer.  I've tried the following: $$\begin{split} (n+1)^2 − (n+1) &amp;= (n+1)(n+1) - (n+1)\\ &amp;=(n+1)(n-1)+2 - (n+1)\\ &amp;=n^2 +n -n -1 +2 -n -1\\ &amp;=n^2 -n, \end{split}$$ but I'm not particularly good at maths so I have no idea if this is even correct, and if it is, what it even means.
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?
6222
I'm working on a double induction problem with the following prompt:     Prove by induction on $n$ that for any real number $q &gt; 1$ and integer $n \ge 0$:   $$ 1 + q + q^2 + q^3 + \cdots + q^{n-1} + q^n = \frac {q^{n+1}-1}{q-1} $$    Based on , I would imagine the solution is:   Base case $(q,n)$ as $(2,n)$ and $(q,0)$ Assume premise for $(2,n)$ and prove $n+1$ Assume premise for $(q,0)$ and prove $q+1$   But even the base cases confuse me...for example on $(q,0)$: $$1+q^0 = \frac {q^{0+1}-1}{q-1} $$ $$2 = 1 $$  That can't right. Maybe I omit the $q^0$ on the LHS, but why?
110000
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.
58589
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.
88073
I am trying to prove the following by Mathematical Induction:  $$\sum_{i=1}^n i\cdot i! = (n+1)!-1\quad\text{for all integers $n\ge 1$}$$   My proof by Induction follows: First prove $P(1)$ is true,  $$\sum_{i=1}^1 i\cdot i! = (1+1)! - 1$$  Then, for all integers $k &gt;= 1$, if $P(k)$ is true then $P(k+1)$ is also true,  $$ \sum_{i=1}^k i\cdot i! = (k+1)! - 1$$ Then, we must show that $P(k+1)$ is true   $$ \sum_{i=1}^{k+1} i\cdot i! = ((k+1)+1)! - 1\\ \sum_{i=1}^{k+1} i\cdot i! = (k+2)! - 1$$  I am currently struggling with the next step to show that $P(k)$ is equal to $P(k+1)$. How would I finish this prove by induction?
99210
Prove that for all  $n \in \mathbb{N}$: $$2^{n-1}(a^n + b^n) \ge (a + b)^n$$ I used induction (for $k = 1,2,...n-1$), and ended up with the following equation:  $$(a + b)(a + b)^{n-1} \le (a + b)(a^{n-1} + b^{n-1})2^{n-2}$$  Now, I have a feeling I'm on the right track, but I'm not totally sure. Could someone give me a pointer one where I should go from here? Any help would be greatly appreciated. Thanks!
88080
I would like to prove by induction the following inequality:  $\frac{4^n}{n+1} &lt; \binom{2n}{n}$, for all natural numbers n > 1.  Any hints?