a^(n+1) - b^(n+1) = (a-b)(a^n+a^(n-1)b+....+ab^(n-1)+b^n) How I do get this result using induction???

If you would have some book in pdf for indicate for me about this subject, I would be very happy.

I have this result for now:
a^(n+1)-b^(n+1) = a^na - b^nb = a^na - a^nb + a^nb - b^nb = a^n(a-b) + (a^n - b^n)b

I have this, but I want to have to get the other equality that is :

a^(n+1) - b^(n+1) = (a-b)(a^n+a^(n-1)b+....+ab^(n-1)+b^n)

But I don't how I can do it this.

1 Answer
Apr 22, 2018

See below.

Explanation:

Let's call this relation we have here p(n):

p(n) : a^(n+1)-b^(n+1) = (a-b)(a^n+a^(n-1)b + ... + ab^(n-1)+b^n)

Let's take the base case to be p(1).

p(1) : a^2 - b^2 = (a-b)(a+b)

nu[p(1)]=1

Note: nu["statement"] is the truth value of that mathematical statement.

Let's assume that our equality is true for some integer x.

p(x): a^(x+1) - b^(x+1) = (a-b)(a^x+a^(x-1)b+...+ab^(x-1)+b^x)

The sum in the second pair of brackets is a bit messy. Let's use Sigma notation to clean it up, which I hope you're familiar with.

p(x): a^(x+1)-b^(x+1) = (a-b)color(blue)(sum_(k=0)^x a^(x-k)b^k)

Since we assumed it to be true, we have

nu[p(x)] = 1

Let's try to see if the case is true for x+1.

p(x+1): a^(x+2) - b^(x+2) =
=(a-b)(a^(x+1)+a^(x)b+a^(x-1)b^2+...+ab^x+b^(x+1))

The result you got is very useful right now:

a^(x+2) - b^(x+2) = a^(x+1)(a-b)+(a^(x+1)-b^(x+1))b

But we know that a^(x+1)-b^(x+1)=(a-b)color(blue)(sum_(k=0)^x a^(x-k)b^k), so we have

a^(x+2) - b^(x+2) = a^(x+1)(a-b) + (a-b)color(blue)(sum_(k=0)^x a^(x-k)b^k)*b

Let's factor out the (a-b).

a^(x+2) - b^(x+2) = (a-b)(a^(x+1)+sum_(k=0)^x a^(x-k)b^(k+color(red)1))

b is raised to the k+1-th power because we had our sum multiplied by b, making each power of it (of b) one degree more.

We should write out the sum we have in brackets:

a^(x+2) - b^(x+2) = (a-b)(a^(x+1)+a^xb + a^(x-1)b^2+...+ab^x+b^(x+1))

This is exactly what p(x+1) is stating! Basically, if p(x) is true, then so is p(x+1). In math terms:

:. nu[p(x)] =1 => nu[p(x+1)]=1

And since the base case is true, we have proven p(n) is true for all integers n bigger than or equal to 1.

In other words,

a^(n+1)-b^(n+1) = (a-b)(a^n+a^(n-1)b + ... + ab^(n-1)+b^n),
forall n in ZZ, n>=1