Can anyone find a pattern?

(For pattern, see a few posts down)

In school we are just learning about the root of quadratics (sum= -b/a, product = c/a etc)
There are many questions we have to do with stuff like: The roots of the equation "..." are A and B. Find the equation with roots 1/A³ and 1/B³
This lead to a lot of stuffing about writing stuff in terms of (A+B) and AB. I then looked for a general formula, and found this:

Aⁿ+Bⁿ =

Code:

---


    k=n
    ___
Sn -\  (-1)k(P)k-1(S)n-2k+2 * [(n-k)C(k-2) + (n-k+1)C(k-1)]
    /__
    k=2

---

where P(roduct) = AB, S(um) = A+B

What i was wondering was:
a) Can anyone proove/disproove this?
b) can (n-k)C(k-2) + (n-k+1)C(k-1) be simplified?

you are still in Algebra 2 or something?

Sorry, but i'm not really sure what algebra 2 is. I am this year doing IB, just started Year 11, and this is in Maths HL.

I know you can easily find it out by finding A and B by
(S +/- sqrt(S^2-4P))/2 and then raise these to the necessary powers, just the formulae i found looked nice, so i thought i'd use it. The combinatorics part at the end is particularly interesting. However, I was just wondering if this can be easily proven?

Hey mate,

I probably have solution to ur problem:

First of all...
^: to the power

Quadratic equation : x^2 - (sum of roots)x + (Product of roots)

(A+B)^3 = A^3 + B^3 + 3AB(A+B)......(i)

U cud get to this just by multipying the square of (A+B) by itself.

now..

From (i),

we can deduce that

A^3 + B^3 = (A + B)^3 - 3AB(A+B).....(ii)

Back to ur orginal roots

1/A^3 and 1/B^3

sum= ( A^3 + B^3 )/(A^3B^3)

or, ((A + B)^3 - 3AB(A+B))/(A^3B^3)

Prod= (1/(A^3B^3))

Now make the quadratic...I am prety sure that this is the way to do it..

well..i know its ooks bit messed up with all power signs...

Hope tis helps..


Sandip

Stop Existing Start Living

I get what u r talking about, but all i was really looking for was someone to proove/disproove that Sum equation in the first post.

Oh, and i am trying it with 3 unknowns now, using A+B+C, AB+BC+CA and ABC to write every Aⁿ+Bⁿ+Cⁿ

However, it may take a while as expanding (A+B+C)ⁿ for any large n takes a while...

Ok, i managed to use rs^a+rt^a+sr^a+st^a+tr^a+ts^a to show that (A+B+C)*P(a) - (AB+BC+CA)*P(a-1) + (ABC)*P(a-2) = P(a+1)
Where P(x) = A^x+B^x+C^x

From this i get:
p(1) = S
p(2) = S² - 2T
p(3) = S³ - 3ST + 3U
p(4) = S⁴ - 4S²T + 4SU + 2T²
p(5) = S⁵ - 5S³T + 5S²U + 5ST² - 5TU
p(6) = S⁶ - 6S⁴T + 6S³U + 9S²T² - 12STU - 2T³ + 3U²

Where
S = A+B+C
T = AB+BC+AC
U = ABC

Now, the pattern at the start is really easy. (powers of S increase by 1, coeff. increase by one)
Even then 2-5-9 infront of the S^xT² is just combinations stuff, but i'm not sure how to explain the # of terms in p(6). Previously, the number of terms in each p() increased by one, but then it increases by two suddenly :confused:. It then only increases by one in p(7). Weird

Anyway, if anyone can see a pattern, post away...