For questions about mathematical induction, a method of mathematical proof. Mathematical induction generally proceeds by proving a statement for some integer, called the *base case*, and then proving that if it holds for one integer then it holds for the next integer. This tag is primarily meant ...

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15
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5answers
77 views

Geometrical interpretation of $(\sum_{k=1}^n k)^2=\sum_{k=1}^n k^3$

Using induction it is straight forward to show $$\left(\sum_{k=1}^n k\right)^2=\sum_{k=1}^n k^3.$$ But is there also a geometrical interpretation that "proves" this fact? By just looking at those ...
3
votes
6answers
83 views

Using induction prove $(n^3)-n$ is divisible by 3 whenever n is a positive number.

I am not sure if I am doing this right, but I have this: There exists an integer $k$. $2k =$ positive number $(2k)^3 - 2k$ [*And this is where I get lost. How does one prove this?]
0
votes
2answers
48 views

how to prove $2^n = {n \choose 0} +{n \choose 1} + \cdots {n \choose n}$ [duplicate]

I have studying my maths book induction chapter and I found things to solve this but I am failed, somebody help me to solve this problem by simple method of mathematical induction. $$2^n = {n \choose ...
0
votes
3answers
17 views

using mathematical induction problem with n variable as exponent

I am a first year Math student and I am looking at problem in my text book which does not have any answers and I have completely no idea how to do this paticular problem. Show, using mathematical ...
1
vote
0answers
35 views

Mathematical induction only dates from the Middle Ages? [migrated]

The technique of "mathematical induction" is a method of proof where you show some theorem is true for some starting integer and prove also that it holding at any arbitrary integer implies it must ...
3
votes
2answers
53 views

Proving that $\sum_{i=2}^n(5i-4)=\frac{n(5n-3)-2}{2}$ for all $n\geq 1$ by mathematical induction

I have this question: Show, using mathematical induction, that for all natural numbers $n$, $$6 + 11 + 16 + 21 + \cdots + (5n-4) = \frac{n(5n-3)-2}{2}$$ I am confused in that that question states ...
1
vote
2answers
16 views

Proving $\sum_{r=1}^n(6r-2)=n(3n+1)$ by induction

A series is defined by $\sum\limits_{r=1}^n(6r-2)$. Use the method of induction to prove that $S_n=n(3n+1)$. I am at the induction step but I am struggling to rearrange $k(3k+1)+6(k+1)-2$ into the ...
2
votes
0answers
18 views

Real-Definite Version of the Axiom of Induction?

I've seen in the answers to a few different questions here on the Mathematics Stack Exchange that one can clearly do mathematical induction over the set $\mathbb{R}$ of all real numbers. I am, ...
2
votes
2answers
166 views

Proof by induction that $x_n>2$ where $x_{n+1}=\frac{2x_n^2 +4x_n -2}{2x_n+3}$

The sequence $x_1$ $x_2$ $x_3$..... is such that $x_1=3$ and $$x_{n+1}=\frac{2x_n^2 +4x_n -2}{2x_n+3}$$ Prove by induction that $x_n>2$ for all $n$. First I proved the base case using $n=1$ as ...
-1
votes
2answers
56 views

Prove $10^{n-1}\le a \lt 10^n$

$$ \forall a \in \mathbb{N}: \quad a = a_{n-1}\times10^{n-1} + a_{n-2}\times10^{n-2} + \dots + a_1\times10 + a_0 \\ a_{n-i} \in \{0;1;2;3;4;5;6;7;8;9\}; \quad a_{n-1} \neq 0 $$ We say that $a$ has ...
5
votes
4answers
2k views

Proof By Induction Divisibility Question: $12\mid 3^n + 7^{n-1} + 8$

Prove that $3^n + 7^{n-1} + 8$ is divisible by $12$ for all positive integers $n$. I have proved it is true for $n=1$ and I have done the 'assume $n=k$' step, but after getting $3^{k+1} + 7^k + 8$, I ...
1
vote
2answers
76 views

How to prove that $9^n - 8n - 1$ is divisible by $64$ for $n\ge 0$?

My textbook provided the following proof: Base case: When $n=0, 9^n-8n-1=0=64\cdot0$, so $64\mid\left(9^n-8n-1\right)$. Induction step: Suppose that $n\in\mathbb N$ and ...
3
votes
8answers
112 views

Proving that $12^n + 2(5^{n-1})$ is a multiple of 7 for $n\geq 1$ by induction

Prove by induction that $12^n + 2(5^{n-1})$ is a multiple of $7$. Here's where I am right now: Assume $n= k $ is correct: $$12^k+2(5^{k-1}) = 7k.$$ Let $n= k+1 $: $$12^{k+1} + 2(5^k)$$ ...
6
votes
7answers
137 views

How to prove that $2^{n+2}+3^{2n+1}$ is divisible by 7 using induction?

I want to prove that $2^{n+2}+3^{2n+1}$ is divisible by 7 using induction. My first step is replace $n$ with $1$. $2^{1+2}+3^{2(1)+1}$ $2^3+3^3$ $8+27$ $35 = 7\times 5$ The next step is assume ...
3
votes
4answers
79 views

Proving that $6^{2n+1} + 1$ is divisible by $7$ for $n\geq 1$ by induction

How should I go about solving a problem like this using induction? Would I: First test $(n = 1)$ so that $6^{2(1)+1} + 1 = 6^3 + 1 = 217/7 = 31$. Then assume $(n = k)$ so that you have $6^{2(k) + 1} ...
6
votes
5answers
92 views

Inductively prove that any natural number $\ge 12$ can be written as the sum of 4s and 5s

I can intuitively see why this is true: Let us assume $n = \alpha \times 4 + \beta \times 5$ with $\alpha,\beta \in \mathbb{N} \cup \{0\}$. $\forall n \in \mathbb{N} \cup \{0\}$: $n \div 4$ will ...
13
votes
7answers
166 views

What are statements about the natural numbers where induction is impossible or unnecessary to prove?

I'm looking for statements like "for all natural numbers, ____" where induction would be impossible or unnecessarily complicated. This is for pedagogical reasons. When students first learn induction, ...
0
votes
2answers
981 views

Number of nodes in binary tree given number of leaves

How would I prove that any binary tree that has n leaves has precisely $2n-1$ nodes ? Given that a binary tree is either a single node "o" or a node with the left and right subtrees contains a binary ...
-1
votes
4answers
96 views

Proof using induction: $15n^2 \leq 2^n$ [on hold]

How to prove this using induction: $15n^2 \leq 2^n$ (with $n \geq 11$) Thanks for your help!
0
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2answers
72 views

Trying to prove $2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})$ and use this to prove… [duplicate]

I am trying to prove this $2( \sqrt{n+1}-\sqrt n )< \frac{1}{\sqrt n}<2( \sqrt{n}-\sqrt {n-1})$ if $n \ge 1$ and using this to prove $2\sqrt{m}-2<\sum^m_{n=1} \frac{1}{\sqrt n}<2( ...
0
votes
2answers
34 views

How to prove the Archimedean property?

The archimedean property states that $$\boxed{~\forall~ ~a,b\in \mathbb{Z}^+~ \exists ~n~|~na\geq b~}$$ I started with disproving .. Suppose $\forall ~\{n,a,b\} \subset \mathbb{Z}^+ , \text{na ...
0
votes
1answer
55 views

Mathematical Induction. Horses made me question my understanding

I recently read about the false inductive proof that all horses are the same colour. There are some mathSE threads about this already (MathSE_thread_1, MathSE_thread_2). After reading this, I now ...
3
votes
3answers
25 views

Having problem in last step on proving by induction $\sum^{2n}_{i=n+1}\frac{1}{i}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i} $ for $n\ge 1$

The question I am asked is to prove by induction $\sum^{2n}_{i=n+1}\frac{1}{i}=\sum^{2n}_{i=1}\frac{(-1)^{1+i}}{i} $ for $n\ge 1$ its easy to prove this holds for $n =1$ that gives ...
3
votes
1answer
32 views

Proof by induction from Spivak's calculus ch 2- 3b

I was cracking my head over the following proof (by induction) from Spivak's calculus. Givens: $ \binom{n+1}{k}=\binom{n}{k-1}+\binom{n}{k} $ and $ n \ge k $ Task: Proof by induction that $ ...
0
votes
1answer
291 views

Induction to prove regular expression

Prove that is if S and T are any regular expressions over the one-letter alphabet, (for example: Σ = {a}), and if n is any natural, then the languages (ST)^n and (S^n)(T^n) are equal. I have to use ...
0
votes
1answer
38 views

Prove using mathematical induction that $n^2 > n+1$ for all $n \ge 2$

I have proved for the initial case $P(2)$ that this is true, but I'm stuck at substituting in $n=k+1$, $(k+1)^2 > (k+1)+1$ = $k^2 + 2k + 1 > k+2$, where do I go from here or have I made a ...
5
votes
1answer
53 views

Partition onto subsets at the same sum

Positive integers $ a_1, a_2,\ldots, a_n $ such that $ a_k\leq k $ and the sum of all these numbers is even and equal to $ 2S $. Prove that the number can be divided into two groups, the amount of ...
1
vote
4answers
3k views

Prove that $ n^3 + 5n$ is divisible by 6 for all $n\in \textbf{N}$ [duplicate]

Prove that $ n^3 + 5n $ is divisible by 6 for all $ n \in \textbf{N} $. I provide my proof below.
4
votes
4answers
74 views

Prove that $1^3 + 2^3 + 3^3 +\cdots+ n^3 = \frac14n^4 + \frac12n^3 + \frac14n^2$

I have to prove that this is true using mathematical induction. I have this: for every $n \in \mathbb N$: $1^3 + 2^3 + 3^3 + ... + n^3 = \frac 14n^4 + \frac 12n^3 + \frac 14n^2$ for $n = 1: 1^3 = ...
1
vote
1answer
47 views

Induction proof of the area of a square

English is not my first language, so I'm sorry if I'm not very clear. I can clarify any question you have. Also, I don't know how to use that math formatting so I apologize for it. I was asked to ...
0
votes
1answer
47 views

Limit of $a_{n+1}= \frac{n}{n+1} a_n$

I think that this sequence $$a_{n+1}= \frac{n}{n+1} a_n$$ can be rewritten as $$a_n= \frac{1}{n+1}a_0.$$ Therefore the limit should be $0$. But my proof by induction turns out wrong. Is my idea ...
2
votes
1answer
37 views

How many Fibonacci Numbers are in the sequence

I have $I_n = \{2^n + 1, 2^n + 2, 2^n + 3, \dots , 2^{n+1}\}$ and I am trying to prove using induction how many Fibonacci numbers are there. First, the length of $I_n$ is $|I_n| = 2^n$ then for $F_0 ...
0
votes
1answer
53 views

How can I prove this statement about square root?

Introduction In computer science there is a field called Formal Methods and Specifications. In this field software designers design softwares by specifying their functionalities in formal methods, ...
0
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0answers
23 views

Demonstration of exponentiation with induction

How can you demonstrate that $a^0 = 1$ and that $a^{-n} = (1/a)^n$ using the principle of mathematical induction?
5
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3answers
80 views

Using induction to prove that $n^2 > n + 1$ for $n\geq2$

Use mathematical induction to prove that $n^2 > n + 1$ for all $n\geq2.$ I have proved that it is true for the initial case $n=2$ as $4>3$, and have assumed the statement to be true for $k^2 ...
1
vote
3answers
93 views

Prove $\frac{1}{n} =\frac{1}{n+1}+\frac{1}{n(n+1)}$ for all integers $n\in\Bbb Z$

I'm pretty sure that we need induction, since it's the format I had to use for previous problems similar to this (it isn't specified that it HAS to be an inductive proof, either, if there is another ...
0
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0answers
57 views

Using induction to prove the “hockey stick theorem”

The question we were given was (where $^nC_c$ is $n$ choose $c$): Show, using induction and the fact that $^nC_c + ^nC_{(c+1)} = ~^{(n+1)}C_{(c+1)}$, the "hockey stick theorem": the sum from $k=c$ ...
0
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0answers
12 views

Prove Ackermann's function by induction

I have to prove the following property $$A(x,y)>x$$ of Ackermann's function. Do we do the following? We will show that $$A(x, y) \geq A(0, x+y)$$ by induction on $k=x+y$. Base case: For $k=0$ ...
0
votes
1answer
22 views

Proving $F_n \ge (\frac{1}{2}(1+\sqrt{5}))^{n-2}$ for $n \in \mathbb{N}_{>1}$ when $F_n$ is the nth Fibonacci number

Let $F_n$ be defined as the nth Fibonacci number. Prove that $F_n \ge (\frac{1}{2}(1+\sqrt{5}))^{n-2}$ with $n \in \mathbb{N}_{>1}$ My approach thus far was to use induction over $n$. ...
30
votes
14answers
4k views

Why doesn't mathematical induction work backwards or with increments other than 1?

From my understanding of my topic, if a statement is true for $n = 1,$ and you assume a statement is true for arbitrary integer $k$ and show that the statement is also true for $k + 1,$ then you prove ...
5
votes
5answers
641 views

Given $n \in \mathbb N$, prove $\sum^n_{k=0}(-1)^k {n \choose k} = 0$

I tried to solve it using induction, but that got me no were, in the middle of the equation stat appearing ks that I don't see how to get out of the equation. I think the easiest way to prove it, it's ...
-1
votes
1answer
33 views

Is there an easier way to prove this induction?

Given that $u_1=1$, $u_{r+1} = \frac{2u_r-1}{3}$ Prove using induction that $u_n = 3(\frac{2}{3})^n-1$ Step 1: prove that $u_1=3(\frac{2}{3})^1-1$ $3(\frac{2}{3})^1-1$ $3(\frac{2}{3}) - 1$ $2-1$ ...
0
votes
1answer
48 views

Show that there exists a unique function with a certain property

I'm trying to prove the following theorem: "Let $~f: \mathbb{N} \times \mathbb{N} \to \mathbb{N}~$ be a function, and let $~c~$ be a natural number. Show that there exists a unique function $~a: ...
1
vote
1answer
42 views

Strong Induction Proof

Prove that $$\sum_{j=1}^n (j)(j+1)(j+2)\cdots(j+k-1) = \frac{n(n+1)(n+2)\cdots(n+k)}{k+1}$$ Hint: $P(n, k)$ is true for all pairs of positive integers $n$ and $k$ if: (a) $P(1, 1)$ is true and $P(n ...
2
votes
2answers
46 views

Prove this binomial identity using induction

prove this identity: $(1-x)^{-k} = \sum\limits_{i>=0} \binom {n+k-1} {k-1} x^n $ using induction. Verification for k=1 is trivial. assuming k= i, proving the identity when k=i+1 is something i ...
0
votes
1answer
24 views

Is this a proof that recursive definition of functions indeed defines a function?

Someone asked me how you prove that defining a function recursively actually defines a function, and then I tried to rigorously prove it. Is it right? Let $\mathbb{N}=\{0,1,2,\dots\}$. For any ...
2
votes
6answers
100 views

Prove that $7$ divides $1 + 2^{(2^n)} + 2^{(2^{n+1})}$ by induction

Prove that $7$ divides $1 + 2^{(2^n)} + 2^{(2^{n+1})}$ by induction. I ran into the above problem. The base case $n=1$ gives $21$ which is divisible by $7$. Now assume it is true for $n$. Then for ...
4
votes
3answers
111 views

Prove by induction that $1+4+7+…+(3n-2) = 2n(3n-1)$

I have an exercise where I, using induction, have to prove the following: \begin{equation*} 1 + 4 + 7 + \ldots + (3n-2) = 2n(3n-1). \end{equation*} I immediately got stuck on the base case with ...
5
votes
8answers
157 views

Prove with induction that $11$ divides $10^{2n}-1$ for all natural numbers.

$$10^{2(k+1)}-1 = 10^{2k+2}-1=10^{2k}\cdot10^{2}-1$$ I feel like there's something in that last part that should make it work, but I can't grasp it. Am I missing something obvious? Am I going in the ...
0
votes
3answers
29 views

Solution check: Let $f_n$ be fibonacci numbers. Prove: $\sum_{k=0}^{n-1} \binom{n+k}{2k+1} = f_{2n-1}$ and $\sum_{k=0}^n \binom{n+k}{2k} = f_{2n}$

The question: Let $f_n$ be fibonacci numbers. Prove: $\sum_{k=0}^{n-1} \binom{n+k}{2k+1} = f_{2n-1}$ and $\sum_{k=0}^n \binom{n+k}{2k} = f_{2n}$ For every $n\in N$. $f_0=f_1=1$, ...