7
$\begingroup$

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

$\endgroup$
  • $\begingroup$ proofwiki.org/wiki/Sum_of_Geometric_Progression $\endgroup$ – lab bhattacharjee Aug 17 '14 at 9:53
  • $\begingroup$ You don't need induction. Just repeat the proof as in the case $n=2$. $\endgroup$ – Quang Hoang Aug 17 '14 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.) $\endgroup$ – Enthusiastic Engineer Aug 17 '14 at 10:11
  • $\begingroup$ @EnthusiasticEngineer It is called Proof by Induction $\endgroup$ – BadAtAlgebra Feb 17 '19 at 7:02
12
$\begingroup$

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 !

$\endgroup$
  • 1
    $\begingroup$ Your sums should end in $\dots+x^2+x$, not $\dots+x+x$. $\endgroup$ – Joonas Ilmavirta Aug 17 '14 at 10:10
  • $\begingroup$ Ahh yes ! will fix it thanks :) $\endgroup$ – AgentS Aug 17 '14 at 10:11
1
$\begingroup$

Use the formula for sum of a geometric series: $$ 1+x+\ldots +x^{n-2}+x^{n-1}=\frac{x^n-1}{x-1} $$

$\endgroup$
0
$\begingroup$

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)

$\endgroup$
-2
$\begingroup$

$(x−1)(x_n−1+x_n−2+...+x+1)=x^n+x^{n-1}+.....+x^2+x-[x^{n-1}+x^{n-2}+....+x+1]=x^n-1$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.