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I found a solution for (c) here on this platform but I'm not sure how to solve rest of the problems because a negative number and a fraction in this situation is a novelty for me which I currently don't know how to deal with. Can somebody help me understand if possible how to solve the remaining examples? Thank you in advance. $$\textrm{(a)} \quad {{-1} \choose {k}} \quad \textrm{for } k \in \mathbb{Z}$$ $$\textrm{(b)} \quad {{-1} \choose {-1-k}} \quad \textrm{for } k \in \mathbb{Z}$$ $$\textrm{(c)} \quad {{1/2} \choose {k}} \quad \textrm{for } k \ge 0$$ $$\textrm{(d)} \quad {{-1/2} \choose {k-1}} \quad \textrm{for } k \in \mathbb{N}$$

Flowers00
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1 Answers1

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$$\textrm{(a)} \quad {{n} \choose {k}} \quad=\quad {{n} \choose {0}}+\quad {{n} \choose {1}}+\cdots+\quad {{n} \choose {k}}\; \textrm{for } k \in \mathbb{Z}$$ Now, even if $n$ is negative or a fraction, we know that

$\quad {{n} \choose {0}}=1$, $\quad {{n} \choose {1}}=n$, $\quad {{n} \choose {2}}=\frac{n\; (n-1)}{2}, \cdots, \quad {{n} \choose {n}}=1$. Just substitute for $n$ in the above fractions to get the result. for example:

$$\textrm{(d)} \quad {{\frac{-1}{2}} \choose {k-1}} \quad=\frac{(\frac{-1}{2})(\frac{-3}{2})\cdots(\frac{3}{2}-k)}{(k-1)!}$$

$$\textrm{(a)} \quad {{-1} \choose {k}} \quad=\frac{(-1)(-2)\cdots(-1+1-k)}{k!}=\frac{(-1)(-2)\cdots(-k)}{k!}$$

Nitish Kumar
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