(Revised for greater clarity)

Show that $$\underbrace{ \left[ \binom {n+1}{n-i} \right] }_ {\mathbf{ v}\; 1\times n}\;\; {\underbrace{\left[\binom 2{1+i-j}\right]}_ {\mathbf M \;n\times n} }^{(\lambda-1)(n+1)} \underbrace{\left[\binom {n+1}{n-i}\right]^T}_ {\mathbf {v^T}\; n\times 1}\ =\sum_{r=0}^{2\lambda}\binom {2\lambda (n+1)}{r(n+1)}(-1)^{r+\lambda}$$ where $i,j=0,1,\cdots (n-1)$; $\lambda>1\ \ (\lambda \in\mathbb Z)$ and $n\geq 2 \ \ (n\in\mathbb Z)$.

This came up in the course of working out this question.

For example, if $n=3, \lambda=2$ the equation above becomes $$\underbrace{ \left[ \binom {4}{3-i} \right] }_ { 1\times 3}\;\; {\underbrace{\left[\binom 2{1+i-j}\right]}_ {3\times 3} }^{4} \underbrace{\left[\binom {4}{3-i}\right]^T}_ {3\times 1}\ =\sum_{r=0}^{4}\binom {16}{4r}(-1)^{r+2}$$ i.e. $$\begin{array} = \big[\binom 43&\binom 42&\binom 41\big] \\ \color{white}{\binom xx} &\hspace{0.1 cm} \\ \color{white}{\binom 33} &\hspace{0.1 cm} \end{array} \left[\begin{array}= \binom 21 &\binom22 &- \\ \binom 20 &\binom 21 &\binom22 \\ - &\binom20 &\binom21 \\ \end{array}\right]^{\;4 }\; \left[ \begin{array} &\binom 43\\ \binom 42\\ \binom 41\\ \end{array} \right]=\scriptsize\binom {16}0-\binom {16}4+\binom {16}8-\binom {16}{12}+\binom {16}{16}$$

Note that, if $\mathbf M$ is an infinite matrix, then $$\begin{align} \mathbf M \mathbf v^T &=\sum_{\text{all }i} \binom 2{1+i-j}\binom 4{3-i}=\left[\binom 6{4-j}\right]\\ \mathbf M^2 \mathbf v^T &=\sum_{\text{all }i} \binom 2{1+i-j}\binom 6{4-i}=\left[\binom 8{5-j}\right]\\ \mathbf {vM}^4\mathbf v^T=\mathbf {vM}^2 \mathbf {M}^2\mathbf v^T &=\sum_{\text{all }i} \binom 8{5-i}^2=\sum_{\text{all }i} \binom 8{5-i}\binom 8{3+i}\\ &=\binom {16}8 \end{align}$$ Hence the truncation of $\mathbf M$ gives rise to the alternating side terms.

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(Original question below - a bit wordy)

Whilst working on this question the following matrix product emerged:

$$P=\underbrace{ \left[ \binom {n+1}{n-i} \right] }_ {\mathbf{ v}\; 1\times n} \underbrace{\left[\binom 2{1+i-j}\right]^{m-2}}_ {\mathbf M \;n\times n} \underbrace{\left[\binom {n+1}{n-i}\right]^T}_ {\mathbf {v^T}\; n\times 1}\qquad\qquad \cdots (1)$$ where $i,j=0,1,\cdots (n-1)$.

By putting $N=n+m-1$, and writing out the product in full for a few values, it appears that

$$\begin{align} P&=\binom {2N}N-2\binom {2N}{N-(n+1)}+2\binom {2N}{N-2(n+1)}+\cdots\\ &=\color{}{\binom {2N}N+2\sum_{r=0}^\infty (-1)^r\binom {2N}{N-r(n+1)}} \end{align}$$ where $\binom ab=0$ for $a<b$.

If $N=\lambda (n+1)$ where $\lambda>1$ and $\lambda=\underbrace{\mu}_\text{integer part}+\underbrace{\delta}_\text{fractional part}$ then $P$ can be written as $$\color{}{P=\sum_{r=0}^{2\mu}\binom {2N}{(\delta+r)(n+1)}(-1)^{r+\mu}}$$ If $\lambda$ is an integer, then $\mu=\lambda, \delta=0$, hence $$\color{} {P=\sum_{r=0}^{2\lambda}\binom {2N}{r(n+1)}(-1)^{r+\lambda} =\sum_{r=0}^{2\lambda}\binom {2(n+m-1)}{r(n+1)}(-1)^{r+\lambda} =\sum_{r=0}^{2\lambda}\binom {2\lambda (n+1)}{r(n+1)}(-1)^{r+\lambda}}\qquad\qquad\cdots (2)\\ $$

Is it possible to derive $(2)$ from $(1)$ directly?

NB - By putting $N=\lambda (n+1)=n+m-1$, we have $m=(\lambda-1)(n+1)+2$, hence $(1)$ becomes

$$P=\left[\binom {n+1}{n-i}\right]\left[\binom 2{1+i-j}\right]^{(\lambda-1)(n+1)}\left[\binom {n+1}{n-i}\right]^T\qquad\qquad \cdots (1)$$ where $i,j=0,1,\cdots (n-1)$.

Note that $$\mathbf M=\left[\binom 2{1+i-j}\right]= \begin{array} & && &i \\ && &0 &1 &2 &3 &4 &\cdots &(n-2) &(n-1)\\ \hline &&j &\color{orange}2 &\color{orange}1 &- &- &- &\cdots &- &-\\ &&1 &\color{orange}1 &\color{orange}2 &\color{orange}1 &- &- &\cdots &- &-\\ &&2 &- &\color{orange}1 &\color{orange}2 &\color{orange}1 &-&\cdots &- &-\\ &&3 &- &- &\color{orange}1 &\color{orange}2 &\color{orange}1&\cdots &- \\ &&\vdots\\ &&(n-2)&- &- &- &- &- &\cdots &\color{orange}2 &\color{orange}1\\ &&(n-1)&- &- &- &- &- &\cdots &\color{orange}1 &\color{orange}2\\ \end{array}$$


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