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Let sequence $\{a_n\}$ be$$2S_n=a^2_{n+1}-a_{n+1},\ \forall n\in N^+$$ where $S_n=a_1+a_2+\cdots+a_n$, and $a_2=a_9$, then the maximum of the $a_1$ is:

A. $3\qquad\qquad$ B. $6\qquad\qquad$ C. $9\qquad\qquad$ D. $12$

I try some $$2S_{n}=a^2_{n+1}-a_{n+1}$$ $$2S_{n-1}=a^2_{n}-a_{n}$$ so $$2a_{n}=2S_{n}-2S_{n-1}=(a^2_{n+1}-a_{n+1})-(a^2_{n}-a_{n})$$ so we have $$(a_{n}+a_{n+1})(a_{n+1}-a_{n}-1)=0$$ so $a_{n+1}=-a_{n}$or $ a_{n+1}-a_{n}=1$ so $a_{2}=-a_{3}=a_{4}=\cdots=-a_{9}$ or $$a_{2}=a_{1}+d,a_{9}=a_{1}+8d$$ this two case I can't get right answer,so How to solve it? Thanks

Maybe I think it's alternate values.such this two condtion $$a_{n+1}=-a_{n},a_{n+1}-a_{n}=1$$ for example $a_{2}=-a_{1},a_{3}-a_{2}=1,a_{4}-a_{3}=1,\cdots $or $$a_{2}-a_{1}=1,a_{3}=-a_{2},\cdots$$ and so on

math110
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  • Not sure if this helps at all, but $2S_{n}=2S_{n-1}+2a_{n}=2\left(a_{n}^{2}-a_{n}\right)+2a_{n}=2a_{n}^{2}\implies S_{n}=a_{n}^{2}$. –  Mar 22 '19 at 03:14
  • @Jake,maybe can help – math110 Mar 22 '19 at 03:17
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    Maybe I am missing something, but if $S_{n}=a_{n}^{2}$, then $a_{1}=S_{1}=a_{1}^{2}$, so $a_{1}^{2}-a_{1}=0\implies a_{1}=0$ or $a_{1}=1$, so the maximum value of $a_{1}$ is $1$? –  Mar 22 '19 at 03:29
  • Does the question state a value of $s_{1}$ or what type of series this is, e.g. arithmetic, geometric, etc.? – NoChance Mar 22 '19 at 03:37
  • @Jake, $2s_{n}=2s_{n-1}+2a_{n}$,this $n\ge 2$ – math110 Mar 22 '19 at 04:06

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