I'm stuck on the induction step where substitution is difficult.
37253
Describe all positive integers,n such that $3^{n+1}$divides $2^{3^n}+1$. I am little confused about what the question asks-if it asks me to find all such positive integers, or if it asks me to prove that for every positive integer n,$3^{n+1}$ divides $2^{3^n}+1$. Kindly clarify this doubt and if it's the former part, please verify my solution-n=1.
96833
How does one prove by induction that  $n! &gt; n^2$  for $n \geq 4$
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.
68740
This one should be pretty easy as its only a first year math problem. it s proof by induction that Im having trouble with.  $$let\;x_{1}=1; for\;each\;n\in \mathbb{N}\;let\;x_{n+1}=\frac{2}{3}x_{n}+1; prove\; x_{n}&lt; 3\;for\;all\;n$$  If anyone is willing to spell it out really obviously step by step and explain how they got from one step to the next I would appreciate it a lot. Thanks.
112005
I tried the methods shown in  but I cannot extend them well into 4 numbers.
104388
The problem  The sides of a circular track contain a sequence of cans of gasoline. The total amount in the cans is sufficient to enable a certain car to make one complete circuit of the track, and it could all fit into the car's gas tank at one time. Use mathematical induction to prove that it is possible to find an initial location for placing the car so that it will be able to traverse the entire track by using the various amounts of gasoline in the cans that it encounters along the way.  My solution  Given any sequence of cans, we know that the sum of all the gasoline will traverse the car around the entire track. What is not known is the initial location of the car.  Basis step: The track has $1$ can, therefore the car will be able to traverse the track from some starting point.  $S(1) =$ some starting point; $S(1) = p_1$, where $p_1$ is the initial starting point.  Induction hypothesis: $S(k) = p_k$, such that $p_k$ is some starting point that, when traversing the track, if at any point on the track the car runs out of fuel, that point will not work. In order to find $p_k$, one will have to exhaust all possible finite points on the track in order to find the initial point that will traverse the car around the track where some nth tank will have enough fuel to get to the next tank.  Thus, $S(k + 1) = p_{k + 1}$, where $p_{k + 1}$ is some point. If at any point in traversing the track the car runs out of fuel, the chosen $p_{k + 1}$ will not work. A car will run out of fuel if it cannot travel from some $n$-th fuel tank to the $(n + 1)$-th fuel tank.  Done.  Is this right?
96688
My question concerns the inductive step of the proof below. I've typed out the work I've done so far. If anyone has any suggestions on how to proceed thanks in advance.  If $n\in\mathbb{N}$, then $(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^n})\ge\frac{1}{4}+\frac{1}{2^{n+1}}$.  $Proof.$ We use mathematical induction.  Base Case. For $n=1$ we see that $1-\frac{1}{2}\ge\frac{1}{4}+\frac{1}{4}$ since $\frac{1}{2}\ge\frac{1}{2}$.  Inductive Step. Let $n&gt;1$. Assume our statement is true for $n$. Now observe that  $\begin{align}(1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n+1}})&amp;\ge\\ \left((1-\frac{1}{2})(1-\frac{1}{4})(1-\frac{1}{8})(1-\frac{1}{16})\cdots(1-\frac{1}{2^{n}})\right)(1-\frac{1}{2^{n+1}})&amp;\ge(\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$  My question is: How do we get from $\begin{align} (\frac{1}{4}+\frac{1}{2^{n+1}})(1-\frac{1}{2^{n+1}}) \end{align}$ to $\frac{1}{4}+\frac{1}{2^{(n+1)+1}}$?  Most of what I've attempted has led to tedious calculations that do not simplify to the desired result. I've also tried making the argument that $(1-\frac{1}{2^{n+1}})$ must be $\ge$ to $\frac{1}{2}$ but then I'm stuck with $\frac{1}{8}+\frac{1}{2^{(n+1)+1}}$... Is my approach incorrect here?
101647
I'm working on proving the following statement using induction:  $$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$  Fair enough. I'll start with the basis step:  Basis Step: (n=1) $$ \sum_{r=1}^n \frac{1}{r^2} \le \frac{2n}{n+1} $$ $$ \frac{1}{1^2} \le \frac{2}{1+1} $$ $$ 1 \le 1  \checkmark  $$  Inductive Step: $$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{(n+1)+1} $$ $$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2(n+1)}{n+2} $$  My goal is to prove $\forall_{n\ge1} s(n) \implies s(n+1) $ or that this inequality holds true for all $n\ge1$.  I'm not quite sure to go from here on the inductive step. I understand that I need to basically work some clever substitution and manipulation into the problem to end up with: $$ \sum_{r=1}^{n+1} \frac{1}{r^2} \le \frac{2n}{n+1} $$  However, I'm not quite sure what needs to done to obtain this after attempting a few times.
74704
I'm new to Math Induction, I don't know how it goes from $𝑘 + 1 = 3𝑥 + 5𝑦 + 1$ into $5𝑦 = 5𝑦 − 5 + 5.$  I'm trying to prove that:   Any whole number of at least 8 cents can be obtained using coins of 3 and 5 cents  Assume: There is a number $𝑘$ for which there are $𝑥$ and $𝑦$ such that $𝑘 = 3𝑥 + 5𝑦$  Here's my working:  $𝑘 = 3𝑥 + 5𝑦$  Therefore $𝑘 + 1 = 3𝑥 + 5𝑦 + 1$  $5𝑦 = 5𝑦 − 5 + 5$  $5𝑦 = 5 𝑦 − 1 + 5$  $𝑘 + 1 = 3𝑥 + 5𝑦 + 1$  $𝑘 + 1 = 3𝑥 + 5(𝑦 − 1) + 5 + 1$  $𝑘 + 1 = 3𝑥 + 5(𝑦 − 1) + 6$  $𝑘 + 1 = 3(𝑥 + 2) + 5(𝑦 − 1)$  $𝒌 + 𝟏 = 𝟑𝒙' + 𝟓𝒚'$
99798
How do you interpret the coefficient of inverse  (lambda) in two step ?
68440
$\displaystyle\sum_{i=0}^n2^i=2^{n+1}-1$.  I don't understand induction so I could use some help.
58051
I'm having some trouble with the following problem:   "A french man is trying to prove that any non empty group of french soccer players satisfies the following:  'if the group has at least one player who is better than Messi, then all the members of the group are better than Messi'.  To prove this, the french man uses induction on the number n of members of the set of soccer players. The statement is obvious if n=1. Let's suppose now that the statement is true for sets of k french players. We will use this to prove that the statement is true for every set of k+1 french players. Let {P1, P2,..., Pk+1,} be a set of k+1 french players. Let's suppose that at least one of them is better than Messi. Without losing generality, we can assume that P1 is better than Messi. Let's now consider the sets {P1, P3,..., Pk+1} and {P1, P2,..., Pk} (each one of these sets contains at least a player who is better than Messi).By the induction hypothesis, it follows that all the players of these sets are better than Messi, i.e., P1,P2,...,Pk+1 are better than Messi.   Where is the error in this argument?"  This is my guess:  We have 2 sets of players: {P1, P3,..., Pk+1} and {P1, P2,..., Pk}. I will call the first one S1 and the second one S2. Now, S1 ∪ S2 contains players that are better than Messi. I'm guessing that, somehow, S1 ∩ S2 is empty and, thus, S1 and S2 are disjoint.  Please help me!  Thanks.
25938
$\dfrac{n}{2} \le \dfrac{1}{1}+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2^n - 1} \le n $  I've Tried for hours but didn't got any striking idea. I don't have any efforts to show rather than induction. Please Try If you 've non-inductive proof.
121306
What does  "Prove by induction" mean? Would you mind giving me an example?
11349
I've tried substitution, replacing $\sec^2x$ with $\tan^2x + 1$, and parts and I just hit dead ends every time... Do you need knowledge of higher-level calculus to solve this?
107978
I am doing mathematical induction. I am stuck with the question below. The left hand side is not getting equal to the right hand side. Please guide me how to do it further.     $1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.   Sol:  $P(n):\ 1\cdot 3 + 2\cdot 4 + 3\cdot 5 + \cdots + n(n+2) = \frac{1}{6}n(n+1)(2n+7)$.  $P(1):\ \frac{1}{6}(2)(9) = \frac{1}{2}(2)(3)$.  $P(1): 3$.  Hence it is true for $n=n_0 = 1$.  Let it be true for $n=k$.  $P(k):\ 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) = \frac{1}{6}k(k+1)(2k+7)$.  We have to prove  $P(k+1):\ 1\cdot 3 + 2\cdot 4 + \cdots + (k+1)(k+3)= \frac{1}{6}(k+1)(k+2)(2k+9)$.  Taking LHS: $$\begin{align*} 1\cdot 3 + 2\cdot 4+\cdots + (k+1)(k+2) &amp;= 1\cdot 3 + 2\cdot 4 + \cdots + k(k+2) + (k+1)(k+3)\\ &amp;= \frac{1}{6}(k+1)(k+2)(2k+9) + (k+1)(k+3)\\ &amp;= \frac{1}{6}(k+1)\left[(k+2)(2k+9) + 6k+18\right]\\ &amp;= \frac{1}{6}(k+1)\left[2k^2 + 13k + 18 + 6k + 18\right]\\ &amp;= \frac{1}{6}(k+1)\left[2k^2 + 19k + 36\right]. \end{align*}$$
71832
I've tried getting everything on the left side and transforming it into something squared so that I can prove it's bigger or equals to 0 but I've been unsuccessful.