# About counting the terms of the expression $x^{m-1} + x^{m-2} +\ldots+ x^0$ given that $x=1$

I want to understand how I can count the terms of the expression $x^{m-1} + x^{m-2} +\ldots+ x^0$ when $x=1$.

The result is $m$, I dont know how to count them formally, any advice would be helpful. I'm desperated, not because it is required to do the above, but how can be done, I need to understand the subject. Sorry for my bad english.

PS: It is related to this limit: $$\lim_{x \to 1} \frac{x^m-1}{x-1} = m$$

I dont want to use L'Hôpital's rule, I just use a simple factorization and a change of a variable.

When you set $x=1$, then the value of each of the terms is $1$! (Note that there are no explicit coefficients). Adding the ones then simply tells you how many ones there are.

• I understand that, but i'm looking more like a way to construct a proof that relates the numbers of terms of the expression "(x^m-1 + x^m-2 +...+ x^0)" with the degree of the polynomial before mentioned for instance (x^m-1 + x^m-2 +...+ x^0) when m=2 and x=1: ((1^(2-1)) + (1^(2-2))) =2 when the degree "m" gets higher then (1+1+1+1+1....+1)=m i wanna relate "m" as the degree of the expression, with the total number of terms of the polynomial, like counting them or something. – Omar Alejandro Reyes Angarita Jan 1 '15 at 22:53
• @OmarAlejandroReyesAngarita: What kind of proof are you looking for? There are plainly $m$ terms in the polynomial, and $m$ is also the degree. I don't see any room for additional proofwork here. – Henning Makholm Jan 1 '15 at 23:03
• Mmmm for stance how do i know there are m terms?. – Omar Alejandro Reyes Angarita Jan 1 '15 at 23:38
• @OmarAlejandroReyesAngarita: There is one term for each natural number between $0$ and $m-1$, inclusive. There are by definition, $m$ such numbers. Therefore there are also $m$ terms. Again, which kind of proof are you looking for. – Henning Makholm Jan 1 '15 at 23:41
• @Omar If the above proof isn't clear enough, induct on $m$. – aes Jan 2 '15 at 0:07

The terms of the polynomial can be indexed by the exponents of $x$ which are $m-1, m-2, \dots 0$ and these are simply $m$ consecutive integers in reverse order. Each term has a different exponent and no exponent is omitted.

There isn't a need to calculate anything or to use limits of any sort. As soon as you write "$x^{m-1} + \cdots + x^0$", you have defined the expression to have $m$ terms. There's nothing to prove.

To evaluate your limit, I recommend a proof by induction on $m$.