Prove $ X^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1) - Math Stack Exchange

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Learn more about Teams Prove $ x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ Ask Question Asked 10 years, 3 months ago Modified 5 months ago Viewed 91k times 11 $\begingroup$

So what I am trying to prove is for any real number x and natural number n, prove $$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$$

I think that to prove this I should use induction, however I am a bit stuck with how to implement my induction hypothesis. My base case is when $n=2$ we have on the left side of the equation $x^2-1$ and on the right side: $(x-1)(x+1)$ which when distributed is $x^2-1$. So my base case holds.

Now I assume that $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ for some $n$. However, this is where I am stuck. Am I trying to show $x^{n+1}-1=(x-1)(x^n + x^{n-1}+x^{n-2}+...+x+1)$? I am still a novice when it comes to these induction proofs. Thanks

• elementary-number-theory
• proof-writing

Share Cite Follow edited Aug 17, 2014 at 9:55 Petite Etincelle 14.9k22 gold badges3333 silver badges6161 bronze badges asked Aug 17, 2014 at 9:52 chettylicechettylice 11311 gold badge11 silver badge44 bronze badges $\endgroup$ 4

• $\begingroup$ proofwiki.org/wiki/Sum_of_Geometric_Progression $\endgroup$ – lab bhattacharjee Commented Aug 17, 2014 at 9:53
• $\begingroup$ You don't need induction. Just repeat the proof as in the case $n=2$. $\endgroup$ – Quang Hoang Commented Aug 17, 2014 at 9:53
• $\begingroup$ Prove it for n=1; then, assuming that it is true for n=k, try to show that it is true for n=k+1. It is easy indeed. (I do not know the name of this proving method.) $\endgroup$ – enthu Commented Aug 17, 2014 at 10:11
• $\begingroup$ @EnthusiasticEngineer It is called Proof by Induction $\endgroup$ – BadAtAlgebra Commented Feb 17, 2019 at 7:02

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Sorted by: Reset to default Highest score (default) Date modified (newest first) Date created (oldest first) 15 $\begingroup$

To conclude your induction proof, just multiply x both sides :

$x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1) $

multiply $x$ both sides :

$\begin{align} \\ x^{n+1}-x &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\ x^{n+1}-1 -(x-1) &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x) \\ x^{n+1}-1 &=(x-1)(x^n+x^{n-1}+x^{n-2}+...+x^2+x)+(x-1) \\ \end{align}$

factor $(x-1)$ and you're done !

Share Cite Follow edited Aug 17, 2014 at 10:11 answered Aug 17, 2014 at 10:03 AgentSAgentS 12.2k33 gold badges3838 silver badges7474 bronze badges $\endgroup$ 2

• 1 $\begingroup$ Your sums should end in $\dots+x^2+x$, not $\dots+x+x$. $\endgroup$ – Joonas Ilmavirta Commented Aug 17, 2014 at 10:10
• $\begingroup$ Ahh yes ! will fix it thanks :) $\endgroup$ – AgentS Commented Aug 17, 2014 at 10:11

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You may just open hooks in right half of expression: $$(x - 1)(x^{n - 1} + x^{n - 2} + \dots + x + 1) = x * (x^{n - 1} + x^{n - 2} + \dots + x + 1) - 1 * (x^{n - 1} + x^{n - 2} + \dots + x + 1) = (x^n + x^{n - 1} + \dots + x^2 + x) - (x^{n - 1} + x^{n - 2} + \dots + x + 1) = x^ n - 1 $$ No problems)

Share Cite Follow answered Apr 28, 2019 at 9:05 Алексей КирилловАлексей Кириллов 1 $\endgroup$ Add a comment |

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