Question

Let x,y \epsilon R. Use mathmatical induction to prove the identity.

x^{n+1}-y^{n+1}=(x-y)(x^{n}+x^{n-1}y+...+xy^{n-1}+y^{n})

Answer 1

Explanation:

We will prove by mathematical induction that, for every natural n,

`(x-y)(x^(n)+x^(n-1)y+...+xy^(n-1)+y^(n))=x^(n+1)-y^(n+1)`

We will prove our base case (when n=1) to be true:

Base case:

`(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)`

Inductive hypothesis:

Given a natural n,

`x^(n+1)-y^(n+1)=(x-y)(x^(n)+x^(n-1)y+...+xy^(n-1)+y^(n))`

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
`y\\eq 0.`

`(x-y)(x^(n+1)+x^(n)y+...+xy^(n)+y^(n+1))=(x-y)y((x^(n+1))/(y)+x^(n)+...+xy^(n-1)+y^(n))=(x-y)y((x^(n+1))/(y))+(x-y)y(x^(n)+...+xy^(n-1)+y^(n))=(x-y)y((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)\\`

With this we have proved our statement to be true for n+1.

In conlusion, for every natural n,

`(x-y)(x^(n)+x^(n-1)y+...+xy^(n-1)+y^(n))=x^(n+1)-y^(n+1)`