In the paper by Masbaum, it was shown that the colored Jones polynomials for a twist knot $K_p$ can be written as

\begin{eqnarray} J_{n}(K_p;q)&=&\sum_{k=0}^{\infty} {\cal C}_{K_p}(k) \frac { \lbrace n-k\rbrace\lbrace n-k+1\rbrace \cdots \lbrace n+k\rbrace} {\lbrace n\rbrace} \cr &=&\sum_{k=0}^{n-1} {\cal C}_{K_p}(k) q^{n k} (q^{-n-1};q^{-1})_k (q^{-n+1};q)_k \ , \end{eqnarray}

where $\def\usc{_}$

\begin{eqnarray} {\cal C}_{K_p}(k) &=&(-1)^{k+1} {q_I}^{k(k+3)/2}\sum_{l=0}^k (-1)^l q^{l(l+1)p} \lbrace 2l+1\rbrace \frac {\lbrace k \rbrace! }{\lbrace k+l+1 \rbrace! \lbrace k-l\rbrace!}\cr &=& (-1)^{k+1} {q}^{k(k+3)/2}\sum_{l=0}^k (-1)^l q^{l(l+1)p+l(l-1)/2}(q^{2l+1}-1) \frac{(q;q)\usc{k}}{(q;q)\usc{k+l+1} (q;q)_{k-l}} \ . \end{eqnarray}

For the trefoil ${\bf 3_1}$, the twist number $p$ is equal to 1. Then the paper says ${\cal C}_{K_1}(k)=(-1)^{k} {q}^{k(k+3)/2}$. My first question: how can it be shown? Namely, does anybody have idea how to prove this? \begin{equation} \sum_{l=0}^k (-1)^l q^{l(l+1)+l(l-1)/2}(q^{2l+1}-1) \frac{(q;q)\usc{k}}{(q;q)\usc{k+l+1} (q;q)_{k-l}}=-1 \end{equation}

In addition, the recent paper (footnote in p.13) showed a simpler form of the colored Jones polynomial for the trefoil \begin{equation} J_n({\bf 3_1};q)=\sum_{k=0}^{n-1} q^{n(k+1)-1} (q^{n-1};q^{-1})_k \ . \end{equation}

My second question: how can one prove that these two expressions are the same? \begin{equation} \sum_{k=0}^{n-1} q^{n(k+1)-1} (q^{n-1};q^{-1})\usc{k} =\sum_{k=0}^{n-1} (-1)^{k} q^{k(k+3)/2} q^{n k} (q^{-n-1};q^{-1})_k (q^{-n+1};q)_k \end{equation}

Although I can check both the identities for $k=2,3$ or $n=2,3$, I have had a hard time to prove them.

Some notations are fixed. \begin{eqnarray} & & \lbrace n\rbrace={q_I}^{n}-{q_I}^{-n}, \; {q_I}^2=q, \\ [n]=\frac {\lbrace n\rbrace}{\lbrace 1\rbrace}, \cr & & \lbrace n\rbrace!=\lbrace n\rbrace\lbrace n-1\rbrace\cdots \lbrace 1\rbrace, \; (x;q)_n=(1-x)(1-x q)\cdots (1-x q^{n-1}) \end{eqnarray}