Homology groups of the pair $(D, \partial D)$ where $D$ is a 2-disc with $k$ open discs removed For my algebraic topology course I am trying to solve an exercise that reads as follows:

Let $D$ be a 2-disc with $k$ open discs removed. Compute the homology of the pair $(D, \partial D)$.

So far I have computed the homology groups of both the boundary $\partial D$ and $D$ itself (see below). Now I wanted to use this result and the exact sequence:
$... \rightarrow H_n(\partial D) \rightarrow H_n(D) \rightarrow H_n(D,\partial D) \rightarrow H_{n-1} (\partial D) \rightarrow ...$
to compute the missing homology groups. However I keep confusing myself in this step and don't know what I have to do as I don't seem to understand the elements of the relative homology enough
This is what I have come up with so far: using a CW structure I computet
$\newcommand{\Z}{\mathbb{Z}}$
$H_0(D) = \Z$
$H_1(D) = \Z^{2k}$
$H_2(D) = \Z$
$H_n(D) = 0$ for $n \geq 3$
and for the boundary
$H_1(\partial D) = \Z^{k+1} = H_0(\partial D)$
$H_n(\partial D) = 0$ for $n \geq 2$
I would be very grateful if someone could  help me with the computations of the relative homology groups now as I don't seem to understand what I have to do next.
 A: First of all I disagree with the computation of $H_{\bullet}(D)$. Using a CW structure I obtain :
$$H_k(D) = 0 \quad \text{if } k\geq 2$$
and $H_0(D) = R,H_1(D) = R^k$.
Using the same CW structure, we can compute the map $H_i(\partial D)\to H_i(D)$ induced by the inclusion $\partial D\hookrightarrow D$. In degree $0$, it is the map $\alpha:R^{k+1}\to R$ sending every member of the canonical  basis to $1$. In degree $1$, it is the map $R^{k+1}\to R^k$ defined by the matrix :
$$M = \left(\begin{array}{ccc|c}1 & &  &1\\
&\ddots & &\vdots \\&&1&1\end{array} \right)$$
Now, the long exact sequence in homology is :
$$0 \to H_2(D,\partial D) \to R^{k+1}\overset{M}\longrightarrow R^{k}\to H_1(D,\partial D) \to R^{k+1}\overset\alpha\longrightarrow R\to H_0(D,\partial D )\to 0$$
This tells us that :
$$H_2(D,\partial D) = \ker(M) = R$$
and
$$H_0(D,\partial D)= \operatorname{coker}(\alpha) = 0$$
as $\alpha$ is surjective.
Lastly, as $M$ is surjective, we get :
$$H_1(D,\partial D) = \ker(\alpha) = R^{k}$$
