Step-by-step explanation:
We will prove by mathematical induction that, for every natural n,
[tex](x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=x^{n+1}-y^{n+1}[/tex]
We will prove our base case (when n=1) to be true:
Base case:
[tex](x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=(x-y)(x^{1}+y^{1})=x^2-y^2=x^{1+1}-y^{1+1}[/tex]
Inductive hypothesis:
Given a natural n,
[tex]x^{n+1}-y^{n+1}=(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})[/tex]
Now, we will assume the inductive hypothesis and then use this assumption, involving n, to prove the statement for n + 1.
Inductive step:
Observe that, for y=0 the conclusion is clear. Then we will assume that [tex]y\neq 0.[/tex]
[tex](x-y)(x^{n+1}+x^{n}y+...+xy^{n}+y^{n+1})=(x-y)y(\frac{x^{n+1}}{y}+x^{n}+...+xy^{n-1}+y^{n})=(x-y)y(\frac{x^{n+1}}{y})+(x-y)y(x^{n}+...+xy^{n-1}+y^{n})=(x-y)y(\frac{x^{n+1}}{y})+y(x^{n+1}-y^{n+1})=(x-y)x^{n+1}+y(x^{n+1}-y^{n+1})=x^{n+2}-yx^{n+1}+yx^{n+1}-y^{n+2}=x^{n+2}-y^{n+2}\\[/tex]
With this we have proved our statement to be true for n+1.
In conlusion, for every natural n,
[tex](x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})=x^{n+1}-y^{n+1}[/tex]
