Questions tagged [induction]

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 for questions about induction over natural numbers but is also appropriate for other kinds of induction such as transfinite, structural, double, backwards, etc.

Mathematical induction is a form of deductive reasoning. Its most common use is induction over well-ordered sets, such as natural numbers or ordinals. While induction can be expanded to class relations which are well-founded, this tag is aimed mostly at questions about induction over natural numbers.

In general use, induction means inference from the particular to the general. This is used in terms such as inductive reasoning, which involves making an inference about the unknown based on some known sample. Mathematical induction is not true induction in this sense, but is rather a form of proof.

Induction over the natural numbers generally proceeds with a base case and an inductive step:

  • First prove the statement for the base case, which is usually $n=0$ or $n=1$.
  • Next, assume that the statement is true for an input $n$, and prove that it is true for the input $n+1$.

The following variant goes without a base case: Assuming the statement is true for all $n\in\mathbb N$ with $n < N$, prove that is true for $N$, too. This has to be done for all $N\in\mathbb N$.

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Question on proof of $1+2+\dots+n=\frac{n(n+1)}{2}$ by induction.

I saw some video where it needs to prove $1+2+\dots+n=\frac{n(n+1)}{2}$ inductively. So it has to be true if $k=1$ and $k+1$ are true. So, for $k=1$: $$1=\frac{1(1+1)}{2}=\frac{1(2)}{2}=\frac{2}{2}=1$$ it is valid. For $k+1$ here is the proof he…
Garmekain
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Show that there exist rational numbers $a_{r1},...,a_{rr}$ such that $\sum_{k=1}^{n} k^{r} = \frac{1}{r+1}n^{r+1}+a_{rr}n^r+...+a_{r1}n$

Let $r$ be a natural number. Show that there exist rational numbers $a_{r1},...,a_{rr}$ such that $$\sum_{k=1}^{n} k^{r} = \frac{1}{r+1}n^{r+1}+a_{rr}n^r+...+a_{r1}n$$ for all natural numbers $n$. I have attempted to prove this statement via…
B.Swan
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Can't find the mistake in my inductive step

Few days ago I was solving some induction execises, and I tried and solved this one. Statement What is wrong with this “proof”? “Theorem” For every positive integer $n$, if $x$ and $y$ are positive integers with $\max(x, y) = n$, then $x = y$. Basis…
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Don't understand an induction solution.

I tried to solve the following induction proof question before giving up and looking at the solution, however the solution didn't really help me either. Question Re-prove the formula for the sum of a geometric series with first term $a$, common…
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Mathematical Induction on Fibonacci numbers

I have ben stuck on this for a while. Let $F(N)$ be the Fibonacci numbers with $F(1)=F(2)=1$. Show that $4(-1)^n + 5(F(N))^2$ is a square for all integers $N$.
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The sum of the fourth powers of the first $n$ positive integers

I am studying mathematical induction and most of the times I have to prove something. Like, for example: $1 + 4 + 9 + ...+ n^2 = \frac{n(n+1)(2n+1)}{6}$ This time I found a question that ask me to find a formula for $1 + 16 + 81 + .... + n^4$ How…
EGS
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How do I prove this using mathematical induction?

\begin{align} 1\cdot3+2\cdot4+3\cdot5+...+n(n+2) = \frac{n(n+1)(2n+7)}{6} \end{align} Using the mathematical induction step I arrive at this : \begin{align} 1\cdot3+2\cdot4+3\cdot5+...+n+1(n+3) = \frac{n+1(n+2)(2n+9)}{6} \end{align} And I don't see…
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Why is this a valid induction rule?

To show a property P is true of all the nonnegative integers, show that P(0) and P(1) are true, that P(n) is true if n is a prime, and for all m, n ∈ N,(P(m) ∧ P(n) → P(mn)). I assume that P(0) and P(1) are base cases... and that P(n) is true if n…
pajkatt
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Can I apply Induction here ??

I have got a question Which is as follow: If there are n participants in a knockout tournament then prove that (n-1) matches will be needed to declare a champion. If I prove this problem using knowledge of combinatorics then it is not so hard, but…
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induction, n + 2

Task is to prove that at start of the N-round, count of hydras heads is divisible by 7. Rules of regrowing hydras heads are (all heads are cut off at start of each round, except black ones): first head is brown (just 1 head), by cutting brown head…
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Prove by induction that $7^{2n}-48n-1$ is divisible by 2304

$$P(n):2304\mid7^{2n}-48n-1$$ I've done the base case; $P(1)$ is true because the expression then evaluates to zero, which is divisible by 2304. Now I'm stuck on the inductive step: proving $P(m+1)$ true if $P(m)$ is true. I do know this…
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Prove by mathematical induction that $1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^n-1} > \frac{n}{2}$

1)Base of induction n=2 : $${1+\frac{1}{2}+\frac{1}{3}>\frac{2}{2}}$$ 2)Assuming right for n=k $$1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2^k-1} > \frac{k}{2}$$ and now need to prove that right for…
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Show that for all Harmonic numbers $H(k)$, the inequality $H(2^k) \leq 1 + k$ holds for all natural numbers.

Hi so the question is prove that for all harmonic numbers $$H(k) = 1 + \frac{1}{2} + \frac13 + \dotsb + \frac{1}{k}$$ the inequality $H(2^k) \leq 1 + k$ holds for all natural numbers. I'm not going to put my entire answer here but was wondering if…
Will
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How to prove $2^{2n}\leq n!$ for all $n\geq 9$ with induction?

So, first step, base of induction is correct. When we take $n=9$, we get $262114\leq 362880$. So, we assume that $2^{2n}\leq n!$ for all $n\geq 9$ is correct and we want to prove that $2^{2n+2}\leq (n+1)!$ is correct for all $n\geq 9$. In…
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Formula that connects b_n and b_{n+1}

I have to use the principle of mathematical induction to show that $6^n - 5n+4$ is divisible by 5 for all integers $n>0$. No problem. I finished this part of the question. The next thing I am asked to do is to make a formula that connects $b_n = 6^n…
Labbiqa
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