Madness continues
( S 1 + S 5 ) ( S 1 + S 9 ) = ( S 7 + S a ) ( S 7 + S b )
For S n = x n + x n 1 , if a and b are constants that satisfy the equation above, find the value of a + b .
This is one of my original Madness problems .
The answer is 4.
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4 solutions
Moderator note:
Very nice! How did you come up with this question? Were you thinking of complex numbers?
No, i didn't think about complex numbers. I accidentally got the idea of using this property. I think one could make even more fascinating questions using this. Thanks!
If you didn't know this property that what is another possible solution that doesn't involve a lot of tedious multiplying?
Where do you get such type of questions
Appreciate your last but one step. Up voted. Just gave my solution to show a different approach.
Did the same way
You know what , I just love this form like anything!!
Yes i can't agree more. The reason for my obsession with this form is its provocative and sheer aesthetic elegance; and the whole realm of formalisms it brings into mathematics. One can't help but be enraptured by the mesmerizing beauty it, so brilliantly, carries.
W L O G L e t i > j , t h e n S i ∗ S j = ( X i + X − i ) ∗ ( X j + X − j ) = { X i + j + X − ( i + j ) } + { X i − j + X − ( i − j ) } = S i + j + S i − j . ( S 1 + S 5 ) ( S 1 + S 9 ) = ( S 7 + S a ) ( S 7 + S b ) , ⟹ S 1 ∗ S 1 + S 1 ∗ S 9 + S 5 ∗ S 1 + S 5 ∗ S 9 = S 7 ∗ S 7 + S 7 ∗ S b + S a ∗ S 7 + S a ∗ S b ∴ f r o m a b o v e S 2 + S 0 + S 1 0 + S 8 + S 6 + S 4 + 1 4 + S 4 = S 1 4 + S 0 + S 7 + b + S 7 − b + S 7 + a + S 7 − a + S a + b + S a − b Equating the powers from left and right 2+0+10+8+6+4+14+4=14+0+(7+b)+(7-b)+(7+a)+(7-a)+(a+b)+(a-b) ⟹ 4 8 = 4 2 + 2 a . ∴ a = 3 , We can see from the above that only one term each on both side has highest power 14. It is self-cancling Next, high powers are 10 and 8, it can be (7+a)=10 and (7+b)=8, OR (7+b)=10 and (7+a)=8. B u t a = 3 . S o b = 1 . ∴ a + b = 3 + 1 = 4
Put x=cos∆+isin∆ use De Moivre's theorem and simplify.... As simple as that!!!
We can solve this problem real fast on applying the beautiful property, S i S j = S i + j + S i − j ⇒ ( S 1 + S 5 ) ( S 1 + S 9 ) = ( S 2 S 3 ) ( S 4 S 5 ) = ( S 2 S 5 ) ( S 3 S 4 ) = ( S 7 + S 3 ) ( S 7 + S 1 ) ⇒ a + b = 3 + 1 = 4