You miss; I hit - 4
If x 2 + x + 1 = 0 , then find n = 1 ∑ 5 0 ( x n + x n 1 ) 3 .
The answer is 94.
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3 solutions
Moderator note:
Good clear explanation of how to simplify the problem step by step.
I accidentally posted the solution without editing and deleted it and the site won't let me post another one, so I post it in a comment- sorry!
I denote P n = x n + 1 / x n . The condition is x 2 + x + 1 = 0 , we divide by x (since x is obviously not zero) and derive that x + 1 / x + 1 = 0 or P 1 = − 1 . We can also prove that P 2 = P 1 2 − 2 , P n + 1 = P 1 P n − P n − 2 . If we follow the sequence { P n } we find out that it follows the next pattern (it becomes periodic and it's easy to prove by induction): P 3 k + 1 = P 3 k + 2 = − 1 , P 3 k = 2 . Cubing the sequence will give us that the sum is actually 1 6 ∗ 6 − 2 = 9 4 .
Oh! That's Great.Nice to hear that.
Congrats! Did it almost the same way.
Congrats man!
Same way as Noam Pirani . Putting it slightly differently:
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We find this is a periodic function with a period of 3. And the following result
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X
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=
8
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n
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∴
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1
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5
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n
+
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1
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3
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1
∑
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n
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3
−
(
x
5
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5
1
1
)
3
=
3
5
1
∗
6
−
8
=
9
4
Yeah I did it the same way...gud ques..
Thanks!!... :)
This is my 5 0 0 t h solution.
Since x 2 + x + 1 = 0 we have x = ω which is complex cube root of unity.Observe that:
S = n = 1 ∑ 5 0 ( x n + x n 1 ) 3 = n = 1 ∑ 5 0 x 3 n + x 3 n 1 + 3 ( x n + x n 1 ) = n = 1 ∑ 5 0 1 + 1 + 3 ( x n + x n 1 ) = 1 0 0 + 3 n = 1 ∑ 5 0 ( x n + x n 1 )
Since ω 2 + ω + 1 = 0 , ω 3 = 1 and ω 1 = ω 2 and ω 2 1 = ω , we can divide the terms of ∑ n = 1 5 0 ( x n + x n 1 ) in groups of three: ( ω + ω 2 + 1 ) = 0 . Since 5 0 ≡ 2 ( m o d 3 ) , the sum of first 48 terms is simply 0 Thus we have:
n = 1 ∑ 5 0 ( x n + x n 1 ) = ω + ω 2 + ω 1 + ω 2 1 = − 1 − 1 = − 2
Thus, we substitute it in main sum:
S = 1 0 0 + 3 ( − 2 ) = 1 0 0 − 6 = 9 4