# Given a condition on $n$ reals

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.

• You post really good questions ^^ – Oussama Boussif Aug 10 '15 at 17:19
• Oh, I am flattered. Thanks. – user167045 Aug 10 '15 at 17:24
• 5050. Gauss' answer ;) – Michael Galuza Aug 10 '15 at 17:35
• @You-know-me he means that $a_n = n$. – Kaster Aug 10 '15 at 17:37
• @You-know-me I don't think it is a guess ;) – Kaster Aug 10 '15 at 17:41

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, $a_i = i$.
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 $$\frac 12 \sum \limits_{i = 1}^k a_i - \frac {k(k - 3)}4 + \sqrt{a_{k+1}-k} = \frac 12 \sum \limits_{i = 1}^{k+1} a_i - \frac {(k+1)(k - 2)}4 \implies \\ \sqrt{a_{k+1} - k} = \frac {a_{k+1}}2 - \frac 14\left [{(k+1)(k-2) - k(k-3)} \right ] = \frac {a_{k+1}}2 - \frac 12(k-1) \implies \\ 4 (a_{k+1} - k) = a_{k+1}^2 - 2a_{k+1}(k-1) + (k-1)^2 \implies\\ a_{k+1}^2 - 2a_{k+1}(k-1+2) + (k-1)^2 + 4k = a_{k+1}^2 - 2a_{k+1}(k+1) + (k+1)^2 = 0 \implies \\ a_{k+1} = k+1$$