  Let $k$ be a field and $\mathcal{C}$ a $k$-linear, Hom-finite triangulated
category with split idempotents. In this paper, we show that under suitable
circumstances, the Grothendieck group of $\mathcal{C}$, denoted
$K_0(\mathcal{C})$, can be expressed as a quotient of the split Grothendieck
group of a higher-cluster tilting subcategory of $\mathcal{C}$.
  Assume that $n\geq 2$ is an even integer, $\mathcal{C}$ is $n$-Calabi Yau and
has an $n$-cluster tilting subcategory $\mathcal{T}$. Then, for every
indecomposable $M$ in $\mathcal{T}$, there is an Auslander-Reiten $(n+2)$-angle
in $\mathcal{T}$ of the form $M\rightarrow T_{n-1}\rightarrow\dots\rightarrow
T_0\rightarrow M$ and \begin{align*}
  K_0(\mathcal{C})\cong K_0^{sp}(\mathcal{T})\big/\big \langle
\sum_{i=0}^{n-1}(-1)^i[T_i]\mid M\in\mathcal{T} \text{ indecomposable }
\big\rangle. \end{align*} Assume now that $d$ is a positive integer and
$\mathcal{C}$ has a $d$-cluster tilting subcategory $\mathcal{S}$ closed under
$d$-suspension. Then $\mathcal{S}$ is a so called $(d+2)$-angulated category
whose Grothendieck group $K_0(\mathcal{S})$ can be defined as a certain
quotient of $K_0^{sp}(\mathcal{S})$. We will show
  \begin{align*}
  K_0(\mathcal{C})\cong K_0(\mathcal{S}).
  \end{align*} Moreover, assume that $n=2d$, that all the above assumptions
hold, and that $\mathcal{T}\subseteq \mathcal{S}$. Then our results can be
combined to express $K_0(\mathcal{S})$ as a quotient of
$K_0^{sp}(\mathcal{T})$.
