 All about flooble | fun stuff | Get a free chatterbox | Free JavaScript | Avatars  perplexus dot info  Summing powers (Posted on 2004-07-01) I thought of three numbers.
Their sum is 6.
The sum of their squares is 8.
The sum of their cubes is 5.
What is the sum of their fourth powers?

 See The Solution Submitted by Federico Kereki Rating: 3.0000 (5 votes) Comments: ( Back to comment list | You must be logged in to post comments.) Complete analytical solution Comment 18 of 18 | Let the numbers be a, b, and c.  Then we have

a + b + c = 6
a2 + b2 + c2 = 8
a3 + b3 + c3 = 5

We will find the (monic) cubic equation whose roots are a, b, and c.
If cubic equation x3 − Ax2 + Bx − C = 0 has roots a, b, c, then, expanding (x − a)(x − b)(x − c), we find

A = a + b + c
B = ab + bc + ca
C = abc

Then B = ab + bc + ca = � [(a + b + c)2 − (a2 + b2 + c2)] = 14.
Hence a, b, c are roots of x3 − 6x2 + 14x − C = 0, and we have

a3 − 6a2 + 14a − C = 0
b3 − 6b2 + 14b − C = 0
c3 − 6c2 + 14c − C = 0

Adding, we have (a3 + b3 + c3) − 6(a2 + b2 + c2) + 14(a + b + c) − 3C = 5 − 6�8 + 14�6 − 3C = 0.
Hence C = 41/3, and x3 − 6x2 + 14x − 41/3 = 0.

Multiplying the polynomial by x, we have x4 − 6x3 + 14x2 − 41x/3 = 0.  Then

a4 − 6a3 + 14a2 − 41a/3 = 0
b4 − 6b3 + 14b2 − 41b/3 = 0
c4 − 6c3 + 14c2 − 41c/3 = 0

Adding, we have (a4 + b4 + c4) − 6(a3 + b3 + c3) + 14(a2 + b2 + c2) − 41(a + b + c)/3 = 0.
Hence a4 + b4 + c4 = 6�5 − 14�8 + (41/3)�6 = 0.

That is, the sum of the fourth powers of the numbers is 0.

Generalization

Using the above approach, we can show that if

a + b + c = r
a2 + b2 + c2 = s
a3 + b3 + c3 = t

then a, b, c are roots of x3 − rx2 + �(r2 − s)x + (�r(3s − r2) − t)/3 = 0.

It then follows that

a4 + b4 + c4  = 4rt/3 − �s(r2 − s) + r2(r2 − 3s)/6. = (r4 − 6r2s + 3s2 + 8rt)/6.

Let f(n) = an + bn + cn, where n is a positive integer.
Then, given the equation x3 − 6x2 + 14x − 41/3 = 0, we can multiply by xn and sum over the three roots to yield the following recurrence relation:

f(n+3) = 6f(n+2) − 14f(n+1) + (41/3)f(n).

Successive applications of this formula allow us to calculate  a5 + b5 + c5,  a6 + b6 + c6, and so on.

Edited on October 28, 2012, 4:06 am
 Posted by Danish Ahmed Khan on 2012-10-24 14:33:51 Please log in:

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