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Using arithmetic series summation formula, I got $S_n=\frac{n(n+1)}{2}$

Let $S_n-k=2001$, $S_{63}=2016$, I got $k=2016-2001=15$

But I think my process is a little loose, what should I do?

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    $\begingroup$ "But I think my process is a little loose" -- I'm not sure what you mean here. What precisely is your question? $\endgroup$ Commented Apr 15, 2023 at 7:08
  • $\begingroup$ If by "a little loose" you mean "not rigorous," then notice that you never show that $63$ is unique. It's not difficult—you can just point out that $n \geq 64$ always means that the difference $S_n-2001$ is greater than $n$—but that's one extra thing you can do. $\endgroup$
    – Brian Tung
    Commented Apr 15, 2023 at 17:19

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I don't know how you chose $S_{63}$, If it is a MCQ exam, your method might be the quickest, nonetheless this is how I would solve it $$2001 + n \ge \frac{n(n+1)}{2} > 2001 \\ 0 \ge n^2-n-4002 \ \ \ \ and \ \ \ \ n^2+n-4002 > 0 $$

Solving this you obtain $n= 63$

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  • $\begingroup$ You are right! I was so stupid that I tried all the $S_n$ one by one. $\endgroup$
    – S.Y.Li
    Commented Apr 15, 2023 at 9:11
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Let $k$ be the missing number. Then, $\frac{n(n+1)}{2} - 2001 = k$.

To tighten your process, you need to convert this lower and upper limit for $n$ and explore all feasible solution in this range. We have

$$ k = \frac{n(n+1)}{2} - 2001 \ge 1 $$

and

$$ k = \frac{n(n+1)}{2} - 2001 \le n $$

Solving these two quadratic inequalities we get $62.7 \le n \le 63.76$. Hence the only feasible value is $n = 63$ and we get $k = 15$.

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  • $\begingroup$ To convince yourself $n=63$ is the only possibility, you can also note that $S_{62}$ is only $1953$, and for lower $n$ it gets only worse, so $n\ge 63$. Then note that $S_{64}$ is $2080$, and the difference to the desired $2001$ is $79$, exceeding the last term $64$ in the sum, and for greater $n$ it just becomes worse, so $n\le 63$. $\endgroup$ Commented Apr 15, 2023 at 7:23

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