Question

Prove that X^(2n) - y^(2n) is divisible by (x+y)? Mathamatical Induction

Answer 1

@apoorvk @RaphaelFilgueiras

Answer 2

PLz read the question Now

Answer 3

Let's factor it! :D
\[x^{2n} - y^{2n} = (x^{n} + y^{n})(x^{n} - y^{n})\]
\[x^{n} - y^{n} = (x - n)(x^{n-1} + x^{n-2}y ......xy^{n-2} + y^{n-1})\]
Therefore:
\[x^{2n} - y^{2n} = (x^{n} + y^{n})(x - y)(x^{n-1} + x^{n-2}y......xy^{n-2} + y^{n-1})\]
And clearly
\[(x - y)\]
is a factor :)

QED

Cheers :D

Answer 4

Oh cr*p, I misread 0.0
Not sure how to prove (x+y) is a factor....
I'll have to pass on that one T.T

Answer 5

Only if n is odd

Answer 6

For n=1
x^[2(1)] - y^[2(1)] = x^2 - y^2 = (x+y)(x-y), which is divisible by x+y
Therefore, the proposition is true for n=1.

Assume that for some positive integers k, x^(2k) - y^(2k) is divisible by x+y, that is x^(2k) - y^(2k) = m (x+y) , where m is a positive integer.

When n = k+1
x^[2(k+1)] - y^[2(k+1)]
= x^2 [x^(2k)] - y^2 [y^(2k)]
= x^2 [m (x+y) +y^(2k)] - y^2 [y^(2k)]
= m(x+y) x^2 + x^2 y^(2k) - y^2 [y^(2k)]
= m(x+y) x^2 + y^(2k) (x^2 - y^2)
= m(x+y) x^2 + y^(2k) (x-y)(x+y)
= (x+y) [ mx^2 + y^(2k) (x-y)] , where [ mx^2 + y^(2k) (x-y)] is a positive integer.
Therefore x^[2(k+1)] - y^[2(k+1)] is divisible by (x+y)
Therefore, the proposition is also true for n=k+1

By principle of mathematical induction, the proposition is true for all positive integers n.

Answer 7

@Callisto
One thing's for sure, I didn't deserve that medal
@Yahoo!
This is the real solution, mine was totally WRONG

Answer 8

@terenzreignz You can't hardly judge if you deserve the medal....