Expansion and simplification of a function f(x) with a binomial of the form (${1-x^n}$)

Question

When I have a function $f(x)=\frac{1-x^n}{1-x}$, how do I express the numerator $(1-x^n)$ as a multiple of the term in the denominator $(1-x)$? i.e., can the numerator be expressed as $(1-x)$$*$$($y$)$

Is that possible? Can binomial expansion be applied here?

Answer 1

hint

For $x\ne 1$, Let

$$S=1+x+x^2+x^3+...+x^{n-1}. $$ then after simplification,

$$S-xS=1-x^n. $$

thus $$S=\frac {1-x^n}{1-x} $$

this is known as the sum of a geometric sequence's terms.

$$1-x^2=(1-x)(1+x) $$ $$1-x^3=(1-x)(1+x+x^2) $$ By induction, $$1-x^{n+1}=1-x^n+x^n (1-x) $$ $$=(1-x)(1+x+x^2+...+x^{n-1})+(1-x)x^n$$ $$=(1-x)(1+x+x^2+...+x^n) .$$

Answer 2

$(1-x)\cdot(1+x+x^2+...+x^{n-1})=1-x^n$ using this property if you cancel $1-x$ $f(x)=1+x+x^2+x^3+\cdots+x^{n-1}$ Does that answer your question?