For all natural numbers n , a sequence { a n } satisfies:
k = 1 ∑ n a k = 2 a n a n 2 + 1
Given that all terms of the sequence are positive, find the value of a 2 0 1 7 1 to five decimal places.
This problem has been modified by me and is a part of <Grade 11 CSAT Mock test> series .
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Very nice.
Let S n = k = 1 ∑ n a k . The first few S n indicate that S n = n . Assuming the claim is true, then we have:
S n 2 a n a n 2 + 1 a n 2 + 1 a n 2 − 2 n a n a n 2 − 2 n a n + n ( a n − n ) 2 ⟹ a n = n = n = 2 n a n = − 1 = n − 1 = n − 1 = n − n − 1 We note that a n < n
Let us prove the claim that a n = n − n − 1 is true for all n ≥ 1 by induction.
Proof:
For n = 1 ,
S 1 2 a 1 a 1 2 + 1 a 1 2 + 1 a 1 2 ⟹ a 1 = a 1 = a 1 = 2 a 1 2 = 1 = 1 Since a n > 0
⟹ a 1 = 1 − 1 − 1 = 1 , therefore the claim is true for n = 1 .
Assuming the claim is true for n , then:
S n + 1 2 a n + 1 a n + 1 2 + 1 a n + 1 2 + 1 a n + 1 2 + 2 n a n + 1 a n + 1 2 + 2 n a n + 1 + n ( a n + 1 + n ) 2 ⟹ a n + 1 = S n + a n + 1 = n + a n + 1 = 2 n a n + 1 + 2 a n + 1 2 = 1 = n + 1 = n + 1 = n + 1 − n + 1 − 1
Therefore, the claim is true for n + 1 and hence true for all n ≥ 1 .
⟹ a 2 0 1 7 1 = 2 0 1 7 − 2 0 1 6 1 ≈ 8 9 . 8 1 0 9 1
(This was the actual solution of the exam)
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Define k = 1 ∑ n a k = S n .
Let a 1 = S 1 = k , then k = 2 k k 2 + 1 , or k 2 = 1 .
Since all terms of this sequence are positive, k = 1 .
Move around some things in the given formula, we get a n 2 − 2 a n S n + 1 = 0 .
Then we can see that a n = S n ± S n 2 − 1 .
It's natural that S n ≥ 1 , and thus S n ≥ S n 2 − 1 and S n 2 − 1 ≥ 0 .
But we know that a n cannot be bigger than S n , so a n = S n − S n 2 − 1 .
Note that S n + 1 = S n + a n + 1 , so plug in a n + 1 = S n + 1 − S n + 1 2 − 1 .
S n + 1 = S n + S n + 1 − S n + 1 2 − 1
S n = S n + 1 2 − 1
S n + 1 2 = S n 2 + 1 , and we see that S n 2 is an arithmetic progression.
S n 2 = n
∴ S n = n .
It is easy to see that a n = S n − S n − 1 = n − n − 1 ( n ≥ 2 ) .
This also holds for n = 1 , as a 1 = 1 − 1 − 1 = 1 .
Therefore, a n = n − n − 1 .
The value we're trying to get, a 2 0 1 7 1 = 2 0 1 7 − 2 0 1 6 1 = 2 0 1 7 + 2 0 1 6 = 8 9 . 8 1 0 9 1 1 7 8 7 0 ⋯ ≈ 8 9 . 8 1 0 9 1 .