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Simplify powers of conjugates? Maths Help

For example, is there an easy way to simplify (1+sqrt(3))^n - (1-sqrt(3))^n ???
Both to the same power, just conjugates of each other and one being subtracted from the other... Can this be simplified?
Reply 1
Original post by ComputerMaths97
For example, is there an easy way to simplify (1+sqrt(3))^n - (1-sqrt(3))^n ???
Both to the same power, just conjugates of each other and one being subtracted from the other... Can this be simplified?

Well if
Unparseable latex formula:

(1+\sqrt{3})^n=1+\sqrt{3} ^n\mathrm{C}_1+\sqrt{3}^2 ^n\mathrm{C}_2 +.... + \sqrt{3}^n


and
Unparseable latex formula:

(1-\sqrt{3})^n=1-\sqrt{3} ^n\mathrm{C}_1+\sqrt{3}^2 ^n\mathrm{C}_2 +.... + \sqrt{3}^n


Then what does that suggest?
Original post by Aph
Well if
Unparseable latex formula:

(1+\sqrt{3})^n=1+\sqrt{3} ^n\mathrm{C}_1+\sqrt{3}^2 ^n\mathrm{C}_2 +.... + \sqrt{3}^n


and
Unparseable latex formula:

(1-\sqrt{3})^n=1-\sqrt{3} ^n\mathrm{C}_1+\sqrt{3}^2 ^n\mathrm{C}_2 +.... + \sqrt{3}^n


Then what does that suggest?


From what you've said, it seems that every odd nth term would double and each even nth term would cancel since we're subtracting the second from the first. This leaves the answer as a multiple of sqrt(3).

Since I posted this question, I've managed to get that:

If
(1+3)n=a+b3(1+\sqrt{3})^n = a + b\sqrt{3}
Then where k = n + 1
(1+3)k=(a+3b)+(a+b)3(1+\sqrt{3})^k = (a+3b) + (a+b)\sqrt{3}

And exact same applied for conjugate, just a minus sign before the sqrt(3) term.

But this doesn't really seem to help me any further.

EDIT : Got an Update - posting it below, gimme a sec to type it out.
(edited 8 years ago)
@Aph

Update:

Got this:

If
(1+3)n(13)n=(a+b3)(ab3)=2b3(1+\sqrt{3})^n - (1-\sqrt{3})^n = (a+b\sqrt{3}) - (a-b\sqrt{3}) = 2b\sqrt{3}
Then where k = n + 1
(1+3)k(13)k=2(a+b)3=2b3+2a3(1+\sqrt{3})^k - (1-\sqrt{3})^k = 2(a+b)\sqrt{3} = 2b\sqrt{3} + 2a\sqrt{3}

So does this mean we have a common difference of 2a32a\sqrt{3}
? Seems like it, now I just need to find a way of evaluating this in more detail.

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