Ideal of rational curve in projective space Let $C_d\subset \mathbb P^d$ be a rational curve of degree $d>2$ ($C_d$ can be reducible) and $n\geq d$. Do we always have $h^0(I_{C_d}(n))=\binom {n+d}{d} - nd-1$?
 A: There is a typo in your binomial coefficient.  What is always true for a Cohen-Macaulay projective curve $C_d$ in $\mathbb{P}^d$ of degree $d$ and arithmetic genus $0$ is the formula for the Hilbert polynomial, $$\chi(\mathbb{P}^d,I_{C_d}(n)) = \binom{n+d}{d} - nd - 1.$$  However, for every $d\geq 4$, there exist smooth, degree $d$, rational curves in $\mathbb{P}^d$ such that for small values of $n$, $h^0(\mathbb{P}^d,I_{C_d}(n))$ is strictly larger than $\chi(\mathbb{P}^n,I_{C_d}(n))$. The first example is when $d$ equals $4$ and $n$ equals $1$.  In general, a degree $d$ rational curve in $\mathbb{P}^d$ that is contained in a hyperplane $\mathbb{P}^{d-1}$ inside $\mathbb{P}^d$ has $h^0(\mathbb{P}^d,I_{C_d}(1))\geq 1$ even though $\chi(\mathbb{P}^d,I_{C_d}(1))$ equals $0$.  
For a specific example, take $d=4$, take $n=1$, take the Segre embedding $\mathbb{P}^1\times \mathbb{P}^1\to \mathbb{P}^3$ with image a smooth quadric surface, and let $C_4$ be the image of a curve of bidegree $(1,3)$ inside $\mathbb{P}^1\times \mathbb{P}^1$.  For a linear embedding $\mathbb{P}^3\subset \mathbb{P}^4$, this degree $4$ curve in $\mathbb{P}^4$ has $$h^0(\mathbb{P}^4,I_{C_4}(1)) = 1 > 0= \chi(\mathbb{P}^4,I_{C_4}(1)).$$
