Show that $\lim_{m\rightarrow\infty}4^{-m}\sum\limits_{j=0}^{\lfloor m/2\rfloor}(-1)^j\binom{m}{j}\binom{3m-4j}{2m-4j}=1$ I wish to show that the following sequence converge to $1$ as $m\rightarrow \infty$
$$\frac{\sum_{j=0}^{\lfloor m/2\rfloor}(-1)^j\binom{m}{j}\binom{3m-4j}{2m-4j}}{4^m}.$$
Any idea of how to do this?
 A: Consider the definition of the Backward Finite Difference
$$
\begin{gathered}
  \nabla _{\,x} f(x) = f(x) - f(x - 1) \hfill \\
  \nabla _{\,x} ^2 f(x) = \nabla _{\,x} \left( {\nabla _{\,x} f(x)} \right) = f(x) - 2f(x - 1) + f(x - 2) \hfill \\
  \quad \quad  \vdots  \hfill \\
  \nabla _{\,x} ^n f(x) = \sum\limits_{0\, \leqslant \,k\, \leqslant \,n} {\left( { - 1} \right)^k \left( \begin{gathered}
  n \hfill \\
  k \hfill \\ 
\end{gathered}  \right)f(x - k)}  \hfill \\ 
\end{gathered} 
$$
It is not difficult to demonstrate that in the case of having a polynomial of degree $m$ we obtain
$$
\nabla _{\,x} ^m p_{\,m} (x) = \nabla _{\,x} ^m (a_{\,m} x^{\,m}  + a_{\,m - 1} x^{\,m - 1}  + \; \cdots \; + a_{\,0})  = m!a_{\,m} 
$$
Now, in your case, because of the limited excursion of the sum index, we have in general that
$$
\sum\limits_{0\, \leqslant \,j\,\left( { \leqslant \,\left\lfloor {m/2} \right\rfloor \, \leqslant \,m} \right)} {\left( { - 1} \right)^j \left( \begin{gathered}
  m \\ 
  j \\ 
\end{gathered}  \right)\left( \begin{gathered}
  3m - 4j \\ 
  2m - 4j \\ 
\end{gathered}  \right)} \quad \left| {\;0 \leqslant m} \right.\quad  = \left. {\nabla _{\,x} ^m \left( \begin{gathered}
  3m + 4x \\ 
  2m + 4x \\ 
\end{gathered}  \right)\;} \right|_{\,x\, = \,0} \quad  \ne \quad \left. {\nabla _{\,x} ^m \left( \begin{gathered}
  3m + 4x \\ 
  m \\ 
\end{gathered}  \right)\;} \right|_{\,x\, = \,0} 
$$
but taking the limit as $m\; \to \,\infty $
$$
\mathop {\lim }\limits_{m\; \to \,\infty } \sum\limits_{0\, \leqslant \,j\,\left( { \leqslant \,\left\lfloor {m/2} \right\rfloor \, \leqslant \,m} \right)} {\left( { - 1} \right)^j \left( \begin{gathered}
  m \\ 
  j \\ 
\end{gathered}  \right)\left( \begin{gathered}
  3m - 4j \\ 
  2m - 4j \\ 
\end{gathered}  \right)} \quad \left| {\;0 \leqslant m} \right.\quad  = \quad \mathop {\lim }\limits_{m\; \to \,\infty } \left. {\nabla _{\,x} ^m \left( \begin{gathered}
  3m + 4x \\ 
  m \\ 
\end{gathered}  \right)\;} \right|_{\,x\, = \,0}  = \mathop {\lim }\limits_{m\; \to \,\infty } 4^{\, m} 
$$
since:
$$
\left( \begin{gathered}
  3m + 4x \\ 
  m \\ 
\end{gathered}  \right) = \frac{1}
{{m!}}\left( {3m + 4x} \right)\left( {3m + 4x - 1} \right) \cdots \left( {3m + 4x - m + 1} \right) = \frac{{4^{\,m} }}
{{m!}}x^{\,m}  +  \cdots 
$$
