Distance Regular Graphs

Question

Given that $\Gamma=(X,R)$ denote a distance-regular graph with diameter $D$ and valency $k=b_0$. How will I show that $k = a_i + b_i + c_i$ for all $0 \leq i \leq D$?

Trying to make use of the definition of distance-regular graphs.

Here is the definition:

Let $\Gamma = (X,R)$ denote a connected graph with diameter $D$. We say $\Gamma$ is distance-regular whenever foar all integers $i$, $0 \leq i \leq D$, and for all $x,y \in X$, the scalars $$c_i = \mid \Gamma_{i-1}(x) \cap \Gamma (y) \mid$$ $$a_i = \mid \Gamma_{i}(x) \cap \Gamma (y) \mid$$ $$b_i = \mid \Gamma_{i+1}(x) \cap \Gamma (y) \mid$$

are constant depending only on $i$ and not on the choice of vertices $x,y$.

Answer 1

A graph $\Gamma=(X,R)$ is distance-regular if it is connected, regular with valency $k$ and diameter $D$, and there exist parameters $b_0=k, \ldots, b_{D-1}, c_1=1, \ldots, c_D$ (the intersection array) such that for all $x \in X, y \in \Gamma_i(x)$, we have that $|\Gamma_{i-1}(x) \cap \Gamma(y)|=c_i, 1 \le i \le d$ and $|\Gamma_{i+1}(x) \cap \Gamma(y) | = b_i, 0 \le i \le D-1$.

Note that in the definition, vertex $y$ must be in $\Gamma_i(x)$. Since the graph is regular with valency $k$, the number of neighbors of $y$ is $k$. Each neighbor of $y$ is in one of the three layers $\Gamma_{i-1}(x), \Gamma_i(x)$ or $\Gamma_{i+1}(x)$ in the distance partition of $\Gamma$ with respect to $x$. Therefore, the sum of the number of neighbors of $y$ in $\Gamma_{i-1}(x), \Gamma_i(x)$ and $\Gamma_{i+1}(x)$ is equal $k$.