On Wed, Nov 24, 1999 at 06:21:48PM -0500, Lucas Wiman wrote:
>you didn't, or if you did (Then please repost it).
The `nice' way of doing it was posted (by somebody else) to the list the
other day, but I'll resend the answer I gave to David Willmore:
===
If (3x^2 - x - 2)^6 = (a_12)x^12 + (a_11)x^11 + ... + (a_1)x + a_0, then
obviously setting x = 1 and calculating would give a_12 + a_11 +
a_10 + ... + a_0. Still with me? OK... Now, if you set x = -1, you get
a_12 - a_11 + a_10 - ... + a_0. So, adding these two together gives
2(a_12) + 2(a_10) + 2(a_8) + ... + 2(a_0), which is exactly twice of
what we want. So, to sum it up:
(3x^2 - x - 2)^6, x = 1 is 3*1 - 1 - 2 = 3 - 1 - 2 = 0.
(3x^2 - x - 2)^6, x = -1 is 3*1 + 1 - 2 = 3 + 1 - 2 = 2.
So, the answer is (1/2)*(0 + 2^6) = 64/2 = 32.
===
I find it quite fascinating that multiple people on the list have been
doing this by brute-force. I tried that too, but when I saw it for the
first time, the other problems left me no time for doing it (so I never
got a correct answer) :-)
I think I got in three or four correct answers for this, but I haven't
really kept count.
/* Steinar */
--
Homepage: http://members.xoom.com/sneeze/
_________________________________________________________________
Unsubscribe & list info -- http://www.scruz.net/~luke/signup.htm
Mersenne Prime FAQ -- http://www.tasam.com/~lrwiman/FAQ-mers
