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\begin{align*}F(t)&=\sum_{k=0}^nS_kt^k=\sum_{k=0}^n[(at)^k+(bt)^k+(ct)^k]=\dfrac{1}{1-at}+\dfrac{1}{1-bt}+\dfrac{1}{1-ct}\end{align*} Since $a,b,c$ are roots of $x^3-px^2+qx-r=0$, hence $at,bt,ct$ are roots of $f(x)=x^3-ptx^2+qt^2x-rt^3=0$. hence $$f(x)=(x-at)(x-bt)(x-ct)$$ then we have $$\dfrac{f'(x)}{f(x)}=\dfrac{1}{x-at}+\dfrac1{x-bt}+\dfrac1{x-ct}$$ \begin{align*}F(t)&=\dfrac{1}{1-at}+\dfrac1{1-bt}+\dfrac1{1-ct}=\dfrac{f'(1)}{f(1)}\\ &=\dfrac{3x^2-2ptx+qt^2}{x^3-ptx^2+qt^2x-rt^3}\bigg|_{x=1}=\dfrac{3-2pt+qt^2}{1-pt+qt^2-rt^3}\end{align*} Restriction : $$|t|<\min(|1/a|,|1/b|,|1/c|)$$
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Sun, 25 Nov 2018 13:23 GMT