Find coefficient of $x^8$ in $(1+x+x^2+x^3+...)^c$
Help me please
I don't know what to do after that because of the exponent $c$
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1Hint: Use $$\dfrac{1}{1 - x} = 1 + x + x^2 + \cdots$$ and the binomial series to answer the question. – Mr Mathster Jan 21 '14 at 02:00
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Hint: You should get $\binom{-c}{8}=\binom{c+7}{8}$. – Macavity Jan 21 '14 at 03:02
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Hint: Note that,
$$ 1+x+x^2 +\dots = \frac{1}{1-x} \implies (1+x+x^2 +\dots)^c = (1-x)^{-c}. $$
Now, use the derivative techniques and evaluate it at zero and divide by $8!$. See a related technique.
Mhenni Benghorbal
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You can do this. First recognize that $$\left(\sum_{n=0}^\infty x^n\right)^c = {1\over(1-x)^c},$$ for $|x| < 1$. Differentiate 8 times and set $x = 0$.
ncmathsadist
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