I am trying to solve the following problem using strong induction, the problem is the following:  For any positive integer $n$, let $T_n$ be the number $1$ if $n&lt;4$ and the number $T_{n − 1} +T_{n − 2} +T_{n − 3}$ if $n\ge 4$. We have $T_1 =1, T_2 =1, T_3 =1, T_4 =T_3 +T_2+T_1=1+1+1=3$, $T_5 =T_4 +T_3 +T_2 =3+1+1=5$, etc. Prove that: $\forall n \in \Bbb{Z}^+$, $T_n &lt;2^n$ $\;$  Sadly, I don't even know where to start this question!
120815
So, here is the question:  For any position integer $n$, let $T(n)$ be the number 1 if $n&lt;4$ and the number $T(n-1) + T(n-2) + T(n-3)$ if $n \geq 4$.  We have $T(1)=1, T(2)=2, T(3)=3$ $$T(4)=T(3)+T(2)+T(1) = 1+1+1+1 = 3$$  $$T(5) = T(4)+T(3)+T(2) = 3+1+1 = 5$$  Prove that: (Universal n is an element of all positive integers), $T(n)&lt;2^n$  Any suggestions?
90457
I have a hard induction problem here that I feel completly lost on.  For every $n\ge3$ there is n different positive integer $a_1,...,a_n$ so that $\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}=1$.  There is only one possible solution for $n=3 $  $\frac{1}{2}+\frac{1}{3}+\frac{1}{6}=1$  for $n=4$ there is 2 possible solutions and they both start with $\frac{1}{2}$.  $\frac{1}{2}+\frac{1}{3}+\frac{1}{8}+\frac{1}{24}$ and $\frac{1}{2}+\frac{1}{4}+\frac{1}{5}+\frac{1}{20}$.  I think the first one is the one.  For $n=5$ there is alot of different possible solutions but it have to start with at least $\frac 13$  I think it have to start with $\frac 12+\frac13$ and then be some kind of sum answer for the last $\frac16$ based on $n$. But I don't know  I have tried finding a pattern but not getting anywhere anymore. Any tips on how to tackle a problem like this?
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
9583
So lets say I have $$ n^{2} \le n! $$  For what positive integers is this not true? $n=2$ and $ 3$  Base case? $$n=4 \implies  16 \le 24 $$  What is the inductive hypothesis and how do I show the inductive proof? Thanks
87479
I'm trying to do homework problems and for the most part I've been getting the results. For this one though, I am having some trouble since its $2^n$ and I can't relate it properly:     Prove using simple induction that $n &lt; 2^n$ for all $n \ge 1$.   So obviously, the basis step holds as $1 &lt; 3$. Now, I assume that $n = k$ holds as well and have to prove for $n=k+1$. This is the step I am having trouble with as I cannot relate the induction hypothesis with what I want to end up with. Can anybody show me a model solution for this one? I think my trouble comes because of the $2^n $
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.
115268
Prove by induction: $$\sum_{k=1}^{n} \binom n k = 2^n -1 $$ for all $n\in \mathbb{N}$.   Today I wrote calculus exam, I had this problem given.  I have the feeling that I will get $0$ points for my solution, because I did this:   Base Case: $n=1$ $$\sum_{k=1}^{1} \binom 1 1 = 1 =  2^1 -1 .$$  Induction Hypothesis: for all $n \in \mathbb{N}$: $$\sum_{k=1}^{n} \binom n k = 2^n -1 $$  Induction Step: $n \rightarrow n+1$ $$\sum_{k=1}^{n+1} \binom {n+1} {k} = \sum_{k=1}^{n} \binom {n+1} {k} + \binom{n+1}{n+1} = 2^{n+1} -1$$  Please show me my mistake because next time is my last chance in this class.
4780
Prove by induction that $\sum\limits_{i=1}^n \frac{1}{n+i} \leq \frac{3}{4}$.   I have to prove this inequality using induction, I proved it for $n=1$ and now I have to prove it for $n+1$ assuming $n$ as hypothesis, but this seems impossible to me because the difference between the sum of $n$ and the sum of $n+1$ is a positive value. Adding a positive value on both sides of the inequality, I don't know how to prove that is always less than or equal to $\frac{3}{4}$.
89373
I cannot complete the third step of induction for this one. The assumption is $3^{2n+3}+40n-27=64k$, and when substituting for $n+1$ I obtain $3^{2n+5}+40n+13=64k$. I've tried factoring the expression, dividing, etc. but I cannot advance from here.
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.
12205
Prove by induction: Find a, and prove the postulate by mathematical induction.  $$\text{For all}~ n &gt; a,~ n! &gt; n^3$$  Where ! refers to factorial.  So far I've done a bit of it, I'll skip right to the inductive statement and assume that $k! &gt; k^3$, then try to prove $(k+1)! &gt; (k+1)^3$  Inductive statement: $(n+1)! &gt; (n+1)^3$  $(n+1)^3= n^3 + 3n^2 + 3n + 1 &lt; n! + 3n^2 + 3n + 1$ (By Induction Hypothesis)  ...But now I'm stuck. Does anyone know where to go next?
66163
I ran across this problem in some old notes, and I frustratingly can't figure out how to do it   Let $a_i$ and $b_i$ be sequences of natural numbers, use induction to show  $\sum_{i=1}^n (a_ib_i)^{1/2} \le (\sum_{i=1}^n a_i)^{1/2}(\sum_{i=1}^n b_i)^{1/2} $  Obviously this is trivial to show for n=1. I can't make much progress on n+1. I've tried various tactics, squaring both sides etc.   Any hint or help would be appreciated. Thanks.
60296
I am a CS undergrad and I'm studying for the finals in college and I saw this question in an exercise list:     Prove, using mathematical induction, that $2^n &gt; n^2$ for all integer n greater than $4$   So I started:  Base case: $n = 5$ (the problem states "$n$ greater than $4$", so let's pick the first integer that matches)  $2^5 &gt; 5^2 \implies 32 &gt; 25$ - ok!  Now, Inductive Step:  $2^{n+1} &gt; (n+1)^2$  now expanding  $2 * 2^n &gt; n^2 + 2n + 1$  Well, I can't go on, I don't know how to finish this proof. Could anyone help me with this one?  Also, I know that the statement (inductive hypothesis) is true. Simply, I'll have a number to an exponent $n$ twice, next to $n$ squared and $2n$ (which could be seen as quadratic and linear functions) which "grows" much lower than the exponential one. It could be easily "proved by example" with any $n$ greater than $4$.  I want to know how to prove it algebraically , so my professor does not freak out.  thanks in advance.
11816
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?
90496
Prove by induction that this number is an integer: $$u_n=(3+\sqrt{5})^n+(3-\sqrt{5})^n$$ Progress I assumed that it holds for $n$ and I tried to do it for $n+1$ but the algebra gets quite messy and I'm unable to prove that the following term is an integer: $\sqrt{5}((3+\sqrt{5})^n-(3-\sqrt{5})^n)$
9458
Proove by the Complete Induction for every $n\in \mathbb{N}, n \geq 1$ $$\frac{1}{\sqrt{1}} + \frac{1}{\sqrt{2}} + \frac{1}{\sqrt{3}} + ... + \frac{1}{\sqrt{n}} &gt; \sqrt{n}$$.  I know only the basis - $\frac{1}{\sqrt{1}} \geq \sqrt{1}$.  But, I dont know how to do the change from the regular induction to the induction of Set theroy(Complete Induction).
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!!!