數學：方程根之和積問題 (1)

Let a = 1/α and b = 1/β.

a and b are distinct roots of ky² + ky + 1 = 0

root (a and b) = [-k ± √(k² - 4k)]/(2k) = [-1 ± √(1 - 4/k)]/2

Then, 1/αⁿ + 1/βⁿ = aⁿ + bⁿ

Any nicer form?

2015-06-22 00:43:22 補充：
The post is due soon.

Do you expect any specific nicer form?

1/αⁿ + 1/βⁿ = { [-1 + √(1 - 4/k)]ⁿ + [-1 - √(1 - 4/k)]ⁿ } / 2ⁿ

2015-06-22 20:48:25 補充：
Oh...
I expect there would be better answers...

2015-06-24 16:46:34 補充：
x² + kx + k = 0, with distinct roots α and β.

1 + k/x + k/x² = 0

k(1/x)² + k(1/x) + 1 = 0

Let a = 1/α and b = 1/β.

Then, a and b are distinct roots of ky² + ky + 1 = 0

roots a and b are [-k ± √(k² - 4k)]/(2k) = [-1 ± √(1 - 4/k)]/2

Then, 1/αⁿ + 1/βⁿ

= aⁿ + bⁿ

= { [-1 + √(1 - 4/k)]ⁿ + [-1 - √(1 - 4/k)]ⁿ } / 2ⁿ

I think the above form is acceptable.