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Let $a_1,a_2 \cdots a_n$ be reals such that $$\sqrt{a_1}+\sqrt{a_2-1}+ \cdots +\sqrt{a_n-(n-1)}=\frac{1}{2}(a_1+a_2+\cdots +a_n)-\frac{n(n-3)}{4}$$ Find the sum of the first $100$ terms of the sequence. I am just unable to think what to do. Pleae give some hints and ideas. Thanks.

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  • $\begingroup$ You post really good questions ^^ $\endgroup$ – Oussama Boussif Aug 10 '15 at 17:19
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    $\begingroup$ Oh, I am flattered. Thanks. $\endgroup$ – user167045 Aug 10 '15 at 17:24
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    $\begingroup$ 5050. Gauss' answer ;) $\endgroup$ – Michael Galuza Aug 10 '15 at 17:35
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    $\begingroup$ @You-know-me he means that $a_n = n$. $\endgroup$ – Kaster Aug 10 '15 at 17:37
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    $\begingroup$ @You-know-me I don't think it is a guess ;) $\endgroup$ – Kaster Aug 10 '15 at 17:41
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Using A.M-G.M inequality $$\sqrt{a_i-(i-1)} \leq \frac{a_i-i+2}{2}$$ Therefore $$\sqrt{a_1}+\sqrt{a_2-1}+ \cdots +\sqrt{a_n-(n-1)}\leq\frac{1}{2}(a_1+a_2+\cdots +a_n)-\frac{n(n-3)}{4}$$

But we know equality in fact holds. So, $a_i = i$.

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    $\begingroup$ And indeed Gauss' answer was right :) $\endgroup$ – Seven Aug 10 '15 at 17:46
  • $\begingroup$ That's really clever! (+1) $\endgroup$ – Hypergeometricx Aug 10 '15 at 18:17
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`Slightly' longer answer :)

I'll skip the base part of the MI. So, it is true for some $k$. $$ \sum \limits_{i = 1}^k \sqrt{a_i - (i - 1)} = \frac 12 \sum \limits_{i = 1}^k a_i - \frac {k(k - 3)}4 $$ Now, let's observe what happens if we take $k + 1$ terms $$ \sum \limits_{i = 1}^{k+1} \sqrt{a_i - (i - 1)} = \frac 12 \sum \limits_{i = 1}^{k+1} a_i - \frac {(k+1)(k - 2)}4 $$ Now, use the equality for the $k$ terms in LHS $$ \frac 12 \sum \limits_{i = 1}^k a_i - \frac {k(k - 3)}4 + \sqrt{a_{k+1}-k} = \frac 12 \sum \limits_{i = 1}^{k+1} a_i - \frac {(k+1)(k - 2)}4 \implies \\ \sqrt{a_{k+1} - k} = \frac {a_{k+1}}2 - \frac 14\left [{(k+1)(k-2) - k(k-3)} \right ] = \frac {a_{k+1}}2 - \frac 12(k-1) \implies \\ 4 (a_{k+1} - k) = a_{k+1}^2 - 2a_{k+1}(k-1) + (k-1)^2 \implies\\ a_{k+1}^2 - 2a_{k+1}(k-1+2) + (k-1)^2 + 4k = a_{k+1}^2 - 2a_{k+1}(k+1) + (k+1)^2 = 0 \implies \\ a_{k+1} = k+1 $$

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