poly question (1 Viewer)

[namburger]

http://i237.photobucket.com/albums/ff3/Namburger/polyquestion1.jpg

Help. Thanks in advance

 

[Affinity]

i. You should be able to do this.
ii. since a^2 - a + 1 = 0, you have a^n - a^(n-1) + a^(n-2) = 0 after multiplying through by a^(n-2). you have something similar with b. now add the 2 equations up and rearrange.

iii. check the first 2 cases, then work on part ii's recurrence formula.. you will probably need the sums to products formulae.

 

[namburger]

i. You should be able to do this.
ii. since a^2 - a + 1 = 0, you have a^n - a^(n-1) + a^(n-2) = 0

umm, can you explain this to me. I can see this case occurs at n = 2, but how come you can assume it occurs for n=3,4, ....

Why are you so smart affinity haha

 

[duy.le]

use the fact that a - 1 =a^2 (where a is the root of the equation)
and like wise b.

just expand the RHS and factorise, when u see a - 1 sub in a^2 and u get the answer, provided u do the same with b. man i hated that question the induction was easy, but dam part ii.

 

[Affinity]

umm, can you explain this to me. I can see this case occurs at n = 2, but how come you can assume it occurs for n=3,4, ....

Why are you so smart affinity haha

There was a comment after that "after multiplying through by a^(n-2). you have something similar with b. now add the 2 equations up and rearrange."
for example if you nultiply that equation by a, you get a^3 - a^2 + a = 0