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While trying to solve an analysis homework problem, I reduced it to this one.

Let $c_1,\ldots,c_n\in\mathbb R\setminus\{0\}$ and $t_1,\ldots,t_n\in(0,1)$, where $t_i$ are distinct. Is it possible that $\forall k\in\mathbb Z_{>0},\sum_{i=1}^nc_it_i^k=0$?

My thoughts are no, but my algebra is pretty insufficient. I considered $\sum_{i=1}^nc_it_i(1+t_i)=\sum_{i=1}^nc_it_i$ (which follows from assuming the proposition) to try to reach contradiction, but my algebra skills stand in the way of getting anywhere.

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    If $n=2$, $t_{1}=t_{2}$ and $c_{1}=-c_{2}$. – 2017-01-26
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    It's unfortunately necessary for the initial problem that I be able to solve this for any $n\in\mathbb Z_{>0}$. – 2017-01-26
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    No, I was giving an example where your $\forall k\in\mathbb{N}_{\geq1},\sum_{i=1}^{n}c_{i}t_{i}^{k}=0$ holds. Since your question says that you are sceptical about this being possible. – 2017-01-26
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    Oh, you missed the part where I said $t_i$ are distinct. It's in the body, I omitted it from the title because it's too long. – 2017-01-26
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    Apologies, I missed that. – 2017-01-26
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    It is not possible. Reference: van Der Monde determinant. – 2017-01-26

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This is not possible:

Since the $t_i$ are distinct, we can assume $0 < t_1 < \dots < t_n < 1$. Now assume that $\sum_i c_i t_i^k = 0$ for all $k$. By factoring out the factor $t_n^k$, this implies

$$ 0 = \sum_{i=1}^{n-1} c_i (t_i / t_n)^k + c_n \xrightarrow[k\to\infty]{} c_n, $$ a contradiction, since $c_n \neq 0$ by assumption. This argument uses that $0 < t_i / t_n < 1$ for $i =1,\dots,n-1$, so that $(t_i/t_n)^k \to 0$ as $k \to \infty$.