Sequentially abnormal sequence!
Suppose the sequence a 1 , a 2 , . . . , a 2 0 1 7 satisfies the following conditions:
a 1 = 0 , ∣ a 2 ∣ = ∣ a 1 + 1 ∣ , . . . ,
∣ a 2 0 1 7 ∣ = ∣ a 2 0 1 6 + 1 ∣ .
Find the minimum value of 2 0 1 7 1 r = 1 ∑ 2 0 1 7 a r
The answer is -0.5.
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1 solution
That i wanted. You have made the right conclusion. Thanks for the solution.
upvoted.
Consider the sequence 0 , − 1 , − 2 , − 3 , ⋯ , − 2 0 1 6 . The sequence satisfies the conditions, and for this sequence the desired sum is − 1 0 0 8 . So this should be the minimum attainable value, right?
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how do you get a 3 = − 2 or a 4 = − 3 ?
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Sorry, the comment was not appropriate. I made a mistake before. The solution is correct.
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@Samrat Mukhopadhyay – don't worry, everybody makes mistakes, I learn making mistakes...
Typo: n 1 r = 1 ∑ 2 0 1 7 a r is actually 2 0 1 7 1 r = 1 ∑ 2 0 1 7 a r .
The minimum value is attained when: { a n = − 1 , when n is an even number a n = 0 , when n is an odd number . Therefore, the minimum value is 2 0 1 7 1 r = 1 ∑ 2 0 1 7 a r = 2 0 1 7 − 1 0 0 8 = − 0 . 4 9 9 7 . . . ≈ − 0 . 5