Minimum No. of vertices required

Question

Prove the following for graph $G$.

1. When length of the shortest cycle in a graph is $k \geq 3$ and the minimum degree of the graph is $d$, then $G$ has minimum $\begin{align*} \\ 1+ \sum_{0 \leq p < \left \lfloor k/2 \right \rfloor} d\cdot (d-1)^p \end{align*}$ vertices for odd $k$.
2. When the length of the shortest cycle in a graph is $k \geq 4$ and the minimum degree of the graph is $d$, then $G$ has minimum $\begin{align*} \\1+ (d-1)^{\left \lfloor k/2 \right \rfloor -1} + \sum_{0 \leq p < \left \lfloor k/2 \right \rfloor-1} d\cdot (d-1)^p \end{align*}$ vertices for even $k$. 

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Combinatorial algorithms: Knuth: categorized under average problem M[23]