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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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### $x^n-y^n$ 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 ...
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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.
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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 ...
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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-... • 2,793 2 votes 4 answers 146 views ### 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 ... • 5,756 0 votes 1 answer 79 views ### 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 ... • 1,227 1 vote 3 answers 133 views ### 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 2 votes 1 answer 57 views ### 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? 1 vote 2 answers 8k views ### 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?
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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. $$\forall(n \geq 0)(4\mid(9^n − 5^n)).$$