Linked Questions
15 questions linked to/from Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$
2
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4
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A more rigorous way to prove this? [duplicate]
I would like to prove the following statement
$$x^n-a^n=(x-a)\sum^{n-1}_{k=0}x^ka^{n-k-1},\qquad\forall n\in\Bbb N_0$$
I can easily prove it by induction using polynomial long division or series ...
- 5,606
1
vote
3
answers
113
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How to prove $ z^n - z^n_0 = (z-z_0) \sum_0^{n-1} z^kz_0^{n-1-k} $ [duplicate]
I want to prove that with $z_0$ a root of $1+z^n$, I have
$$ z^n - z^n_0 = (z-z_0)\sum_0^{n-1} z^kz_0^{n-1-k}$$
- 393
0
votes
2
answers
166
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Induction Proof that $ x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1}).$ [duplicate]
I seek an inductive proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\cdots+xy^{n-2}+y^{n-1}).$ I am stuck.
- 139
0
votes
3
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90
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Prove by induction: $b^n-a^n=(b-a)\cdot \sum_{k=0}^{n-1}{a^k\cdot b^{n-1-k}}$ $\forall n\in \mathbb{N}$ and $a,b\in \mathbb{R}$ [duplicate]
Prove by induction: $b^n-a^n=(b-a)\cdot \sum_{k=0}^{n-1}{a^k\cdot b^{n-1-k}}$ $\forall n\in \mathbb{N}$ and $a,b\in \mathbb{R}$
$\exists n\in \mathbb{N}:b^n-a^n=(b-a)\cdot \sum_{k=0}^{n-1}{a^k\cdot b^...
user666536
1
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2
answers
7k
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Prove that $5^n - 2^n$ is divisible by $3$ for all nonnegative integers $n$ using mathematical induction [duplicate]
Using mathematical induction, prove for all integers n 1 that $5^n - 2^n$ is divisible by 3.
Can someone help me with this?
- 643
2
votes
5
answers
6k
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How do I prove that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\dotsb+xy^{n-2}+y^{n-1})$
I am unsuccessfully attempting a problem from Spivak's popular book 'Calculus' 3rd edition. The problem requires proof for the following equation:
$$x^n-y^n=(x-y)(x^{n-1}+x^{n-2}+\dotsb+xy^{n-2}+y^{n-...
- 23
1
vote
1
answer
3k
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The identity $a^n-b^n=(a-b) (\sum_{i=0}^{n-1}a^ib^{n-1-i})$ [duplicate]
How do I use finite induction to prove that
$$a^n-b^n=(a-b) (\sum_{i=0}^{n-1}a^ib^{n-1-i}), \forall a,b\in \Bbb{R}\space \text{and} \space \forall n \in \Bbb{N}?$$
Ok, for $n=2$ it's fine. $a^2-b^2=(a-...
- 2,488
3
votes
2
answers
1k
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Is there a name for a binomial expansion without coefficients?
I am investigating a problem from George E. Andrews Number Theory (Dover, 1971), discussed previously here:
Induction Proof that $x^n-y^n=(x-y)(x^{n-1}+x^{n-2}y+\ldots+xy^{n-2}+y^{n-1})$
I was led ...
- 158
5
votes
1
answer
417
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Decomposition of $x^n-y^n$
As a part of textbook assignment I was asked to prove that $x^2-y^2=(x+y)(x-y)$, and I did so as follows:
$$x^2-y^2=x^2-y^2+xy-xy=x(x+y)-y(x+y)=(x+y)(x-y)$$
Later, I used similar method to decompose $...
- 221
3
votes
6
answers
168
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Proof that: $(a-b)\cdot\Bigg(\sum_{k=0}^{n}a^{n-k}b^{k}\Bigg)=a^{n+1}-b^{n+1}\text{ }\forall n\in\mathbb{N_{0}}$
I'm trying to prove a more general version of the 3rd binomial equation via mathematical induction which will help me complete another proof.
$$(a-b)\cdot\Bigg(\sum_{k=0}^{n}a^{n-k}b^{k}\Bigg)=a^{n+1}...
- 1,291
0
votes
3
answers
111
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How can I use induction solve this?
How can I show/solve this? I've tried by using the basis step and the inductive step, but just can't seem to get it right.
$$\forall(n \geq 0)(4\mid(9^n − 5^n)).$$
- 19
2
votes
1
answer
55
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For $a \gt b \gt 0$ prove that $a^n – b^n \geq n(a-b) (ab)^{\frac{n-1}{2}}$
For $a \gt b \gt 0$ prove that $a^n – b^n \geq n(a-b) (ab)^{\frac{n-1}{2}}$
What is an appropriate way to solve this?
- 21
0
votes
1
answer
79
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Proof by induction valid or not?
Prove by induction the following:
$$\sum_{i=0}^n x^i = \frac{1-x^{n+1}}{1-x}$$
We want:
$$x^0+x^1+ \ldots + x^n = \frac{1-x^{n+1}}{1-x}$$
I try this for $i=1$ and it works, so I have an initial ...
- 1,157
0
votes
1
answer
99
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$x^n-y^n$ equivalence
I have questions about a formula that is :
$x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \ldots + xy^{n-2} + y^{n-1})$
That's how it's written on my textbook but it seems like I've trouble understanding it ...
- 85
5
votes
0
answers
71
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Verify proof: $a^{n}-b^{n} = (a-b) \sum\limits_{k=0}^{n-1} a^{k}b^{n-1-k}$
A short disclaimer: I do know this question has been asked multiple times here and several answers (including combinatorics) have been given already. However, among all these posts, I did not find ...
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