#16 of March 2016 Grade 11 - Liberal Arts CSAT (Korean SAT) Mock test

Algebra Level 4

For all natural numbers n , n, a sequence { a n } \{a_n\} satisfies:

k = 1 n a k = a n 2 + 1 2 a n \large \sum_{k=1}^n a_k = \frac{{a_n}^2+1}{2a_n}

Given that all terms of the sequence are positive, find the value of 1 a 2017 \dfrac{1}{a_{2017}} to five decimal places.


This problem has been modified by me and is a part of <Grade 11 CSAT Mock test> series .


The answer is 89.81091.

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2 solutions

Boi (보이)
Sep 2, 2017

Define k = 1 n a k = S n . \displaystyle \sum_{k=1}^n a_k=S_n.

Let a 1 = S 1 = k , a_1=S_1=k, then k = k 2 + 1 2 k , k=\dfrac{k^2+1}{2k}, or k 2 = 1. k^2=1.

Since all terms of this sequence are positive, k = 1. k=1.


Move around some things in the given formula, we get a n 2 2 a n S n + 1 = 0. {a_n}^2-2a_nS_n+1=0.

Then we can see that a n = S n ± S n 2 1 . a_n=S_n\pm \sqrt{{S_n}^2-1}.

It's natural that S n 1 , S_n\ge1, and thus S n S n 2 1 S_n\ge\sqrt{{S_n}^2-1} and S n 2 1 0. \sqrt{{S_n}^2-1}\ge0.

But we know that a n a_n cannot be bigger than S n , S_n, so a n = S n S n 2 1 . a_n=S_n-\sqrt{{S_n}^2-1}.


Note that S n + 1 = S n + a n + 1 , S_{n+1}=S_n+a_{n+1}, so plug in a n + 1 = S n + 1 S n + 1 2 1 . a_{n+1}=S_{n+1}-\sqrt{{S_{n+1}}^2-1}.

S n + 1 = S n + S n + 1 S n + 1 2 1 S_{n+1}=S_n+S_{n+1}-\sqrt{{S_{n+1}}^2-1}

S n = S n + 1 2 1 S_n=\sqrt{{S_{n+1}}^2-1}

S n + 1 2 = S n 2 + 1 , {S_{n+1}}^2={S_n}^2+1, and we see that S n 2 {S_n}^2 is an arithmetic progression.

S n 2 = n {S_n}^2=n

S n = n . \therefore~S_n=\sqrt{n}.


It is easy to see that a n = S n S n 1 = n n 1 ( n 2 ) . a_n=S_n-S_{n-1}=\sqrt{n}-\sqrt{n-1}~(n\ge2).

This also holds for n = 1 , n=1, as a 1 = 1 1 1 = 1. a_1=\sqrt{1}-\sqrt{1-1}=1.

Therefore, a n = n n 1 . \boxed{a_n=\sqrt{n}-\sqrt{n-1}}.

The value we're trying to get, 1 a 2017 = 1 2017 2016 = 2017 + 2016 = 89.8109117870 89.81091 . \dfrac{1}{a_{2017}}=\dfrac{1}{\sqrt{2017}-\sqrt{2016}}=\sqrt{2017}+\sqrt{2016}=89.8109117870\cdots \approx\boxed{89.81091}.

Very nice.

Anandh Rajan - 3 years, 5 months ago

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Thank you! >w>

Boi (보이) - 3 years, 5 months ago

Let S n = k = 1 n a k \displaystyle S_n = \sum_{k=1}^n a_k . The first few S n S_n indicate that S n = n S_n = \sqrt n . Assuming the claim is true, then we have:

S n = n a n 2 + 1 2 a n = n a n 2 + 1 = 2 n a n a n 2 2 n a n = 1 a n 2 2 n a n + n = n 1 ( a n n ) 2 = n 1 a n = n n 1 We note that a n < n \begin{aligned} S_n & = \sqrt n \\ \frac {a_n^2 + 1}{2a_n} & = \sqrt n \\ a_n^2 + 1 & = 2 \sqrt n a_n \\ a_n^2 - 2 \sqrt n a_n & = - 1 \\ a_n^2 - 2 \sqrt n a_n + {\color{#3D99F6}n} & = {\color{#3D99F6}n} - 1 \\ \left(a_n - \sqrt n\right)^2 & = n - 1 \\ \implies a_n & = \sqrt n - \sqrt{n - 1} & \small \color{#3D99F6} \text{We note that }a_n < \sqrt n \end{aligned}

Let us prove the claim that a n = n n 1 a_n = \sqrt n - \sqrt{n - 1} is true for all n 1 n\ge 1 by induction.

Proof:

For n = 1 n=1 ,

S 1 = a 1 a 1 2 + 1 2 a 1 = a 1 a 1 2 + 1 = 2 a 1 2 a 1 2 = 1 a 1 = 1 Since a n > 0 \begin{aligned} S_1 & = a_1 \\ \frac {a_1^2 + 1}{2a_1} & = a_1 \\ a_1^2 + 1 & = 2a_1^2 \\ a_1^2 & = 1 \\ \implies a_1 & = 1 & \small \color{#3D99F6} \text{Since }a_n > 0 \end{aligned}

a 1 = 1 1 1 = 1 \implies a_1 = \sqrt 1 - \sqrt{1-1} = 1 , therefore the claim is true for n = 1 n=1 .

Assuming the claim is true for n n , then:

S n + 1 = S n + a n + 1 a n + 1 2 + 1 2 a n + 1 = n + a n + 1 a n + 1 2 + 1 = 2 n a n + 1 + 2 a n + 1 2 a n + 1 2 + 2 n a n + 1 = 1 a n + 1 2 + 2 n a n + 1 + n = n + 1 ( a n + 1 + n ) 2 = n + 1 a n + 1 = n + 1 n + 1 1 \begin{aligned} S_{n+1} & = {\color{#3D99F6}S_n} + a_{n+1} \\ \frac {a_{n+1}^2 + 1}{2a_{n+1}} & = {\color{#3D99F6}\sqrt n} + a_{n+1} \\ a_{n+1}^2 + 1 & = 2\sqrt n a_{n+1} + 2 a_{n+1}^2 \\ a_{n+1}^2 + 2\sqrt n a_{n+1} & = 1 \\ a_{n+1}^2 + 2\sqrt n a_{n+1} + {\color{#3D99F6}n} & = {\color{#3D99F6}n} + 1 \\ \left(a_{n+1} + \sqrt n\right)^2 & = n + 1 \\ \implies a_{\color{#D61F06}n+1} & = \sqrt {\color{#D61F06}n+1} - \sqrt {{\color{#D61F06}n+1}-1} \end{aligned}

Therefore, the claim is true for n + 1 n+1 and hence true for all n 1 n \ge 1 .

1 a 2017 = 1 2017 2016 89.81091 \implies \dfrac 1{a_{2017}} = \dfrac 1{\sqrt{2017}-\sqrt{2016}} \approx \boxed{89.81091}

(This was the actual solution of the exam)

Boi (보이) - 3 years, 9 months ago

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