I'm wondering why and how you get some of these steps in an Induction proof.

Okay, so the question is to prove the following statement by induction:

$1 - x + x^2 - x^3$ $+ ... +$ $x^{2n-2} = \frac{x^{2n-1}+1}{x+1}$

The first step that sort of confuses me is having a base case of $n=2$. Why would you start out with that, and not $n=1$? Maybe it's just trivial to show the $n=1$ base case.

Thes next step that also confuses me would be when we show that this is true for $n = k+1$

We write it out as:

$1 - x + x^2 - x^3$ $+ ... +$ $x^{2k-2} - x^{2k-1} + x^{2k} = \frac{x^{2k-1}+1}{x+1} - x^{2k-1} + x^{2k}$

I don't understand how two terms were added ($- x^{2k-1} + x^{2k}$)

Any sort of explanation of clarification would be helpful! Thanks.

  • $\begingroup$ Technically, there can be as many base cases as wanted. So I believe it was more out of observing that the case $n=1$ is trivial. $\endgroup$ – Matt Groff Oct 29 '12 at 1:23
  • $\begingroup$ Remark $\ $ It's simpler to prove more generally $\rm\ 1 + z + \,\cdots + z^k = \dfrac{1-z^{k+1}}{1-z}.\:$ Yours is the special case $\rm\:z = -x,\ k = 2n-2.$ $\endgroup$ – Bill Dubuque Oct 29 '12 at 2:46
  • $\begingroup$ My guess is that if the first step was to test the base case of $n=2$, then the question asked you to prove it for $n\ge2$. $\endgroup$ – Mike Oct 29 '12 at 4:28

Let $P(n)$ be the statement


you want to prove that $P(n)$ is true for all $n\ge 2$. (I’ll come back to the choice of $2$ later.) For the induction step you assume $P(k)$ for some $k\ge 2$ and try to prove $P(k+1)$. Let’s see what those statements really are. $P(k)$ is


obtained by substituting $k$ for $n$ in $(1)$. $P(k+1)$ is obtained similarly, by substituting $k+1$ for $n$ in $(1)$, so it’s


This can be simplified. First, the righthand side is clearly $$\frac{x^{2k+1}+1}{x+1}\;.$$ The lefthand side is $1-x+x^2-x^3+\ldots+x^{2k}$; if you display a couple more terms, working backwards from the end, you’ll see that it’s


which is $$\Big(1-x+x^2-x^3+\ldots+x^{2k-2}\Big)-x^{2k-1}+x^{2k}\;.$$ Thus, $P(k+1)$ is the statement


which can be usefully parenthesized as


This is useful because $P(k)$, the induction hypothesis, lets us replace


by $\dfrac{2^{2k-1}+1}{x+1}$ in $(2)$ to see that $P(k+1)$ is equivalent to the claim that


which is easily verified by a little elementary algebra.

I don’t know why the induction was started at $n=2$; presumably the theorem was stated that way, as the assertion that $P(n)$ holds for every $n\ge 2$. It could just as well have been started at $n=1$: since $2\cdot1-2=0$, $P(1)$ is


which is certainly true.


Substitute $n=k+1$. Then the last member of the sum is $+x^{2n-2}=+x^{2k}$, and before it is $-x^{2k-1}$.

If $n$ increases by one, we get 2 more summands...

About the first step, yes, $n=1$ is enough, and trivial. But, usually it's good to see for small $n$'s what we are talking about..


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