Prove by mathematical induction that x^2n - y^2n has a factor of x+y.
My answer is incomplete since i do not know what to do next. here it is:
i.) for n=1
x^2(1) - y^2(1) = x^2 - y^2
= (x+y)(x-y)
ii.) Assume that the proposition is true for all n=k, that is, x^2k - y^2k has a factor of x+y for all k¡Ý1.
iii.) n=k+1
x^2(k+1) - y^2(k+1)
=x^2k+2 - y^2k+2
=x^2k ¡ñ x^2 - y^2k ¡ñ y^2
=(tthis part I don't know)
I hope you can help me. Thanks!
Certainly! Let's continue with the proof of mathematical induction.
iii.) n = k+1
We will start with the left-hand side of the equation:
x^(2(k+1)) - y^(2(k+1))
Now, we can use the assumption from step ii., which states that x^(2k) - y^(2k) has a factor of x+y. Therefore, we can write:
x^(2k) - y^(2k) = (x+y) * P(x, y)
where P(x, y) is some expression involving x and y.
Now, let's substitute this expression into our equation:
x^(2(k+1)) - y^(2(k+1)) = (x+y) * P(x, y) * x^2 - (x+y) * P(x, y) * y^2
Next, we can factor out (x+y) from both terms:
x^(2(k+1)) - y^(2(k+1)) = (x+y) * P(x, y) * (x^2 - y^2)
Now, we notice that the term (x^2 - y^2) is the difference of squares, which can be factored as:
x^2 - y^2 = (x+y)(x-y)
So, substituting this back into our equation, we get:
x^(2(k+1)) - y^(2(k+1)) = (x+y) * P(x, y) * (x+y)(x-y)
Now, we can combine the first and last terms:
x^(2(k+1)) - y^(2(k+1)) = (x+y)^2 * P(x, y) * (x-y)
Notice that we have (x+y) as a factor. Therefore, x^(2(k+1)) - y^(2(k+1)) has a factor of (x+y).
Since the statement is true for n = k+1, and we have proven that if the statement is true for n = k, then it is true for n = k+1, we can conclude that the statement holds for all positive integers n by the principle of mathematical induction.
I hope this explanation helps! Let me know if you have any further questions.