How prove this identity wih the sum equal to other sum

Question:

Given $$l\in \mathbb{N^+}$$. $$a_1,\cdots,a_l,b_1,\cdots,b_l$$ are real numbers.$$a_0=b_0=a_{l+1}=b_{l+1}=0$$.Define $$g(m,l)=-\dfrac{\displaystyle\prod _{r=0} ^{l}{(a_m-a_r+b_{r+1}-b_m)}}{\displaystyle\prod _{r=1,r\ne m} ^{l}{(a_m-a_r+b_r-b_m)}}$$ Prove: $$\sum_{m=1}^{l}{g(m,l)}=\sum_{t=1}^{l}{(a_t-a_{t-1})b_t}$$

For example,for $$l=1$$,$$g(1,1)=-\frac{a_1(-b_1)}{1}=(a_1-a_0)b_1$$

Another example:for $$l=2$$, $$g(1,2)+g(2,2)=-\frac{a_1(b_2-b_1)(a_1-a_2-b_1)}{a_1-a_2+b_2-b_1}+\frac{b_2(a_2-a_1)(a_2+b_1-b_2)}{a_2-a_1+b_1-b_2}$$ $$=\frac{(a_2-a_1+b_1-b_2)(a_1 b_1+a_2 b_2-a_1 b_2)}{a_2-a_1+b_1-b_2}=\sum_{t=1}^{2}{(a_t-a_{t-1})b_t}$$

But How to prove general case,Maybe use Lagrange interpolation formula?

Fix $$l$$ (so we may then drop the $$l$$ from $$g(m,l)$$), and let $$\alpha_{i,j}=a_{i}-a_j,\beta_{i,j}=b_i-b_j$$, $$d_{ij}=\alpha_{i,j}-\beta_{i,j}$$. Denote by $$S$$ the main sum $$S=\sum_{m=0}^l g(m)$$.

Now, let us temporarily fix two indices $$i$$ and $$j$$. One can write

$$g(i)=\frac{1}{d_{ij}}\frac{A(a_i-b_i)}{B(a_i-b_i)} \tag{1}$$

where $$A(X)=\prod_{r=0}^{l} (X+b_{r+1}-a_r)$$ is independent of $$i$$ or $$j$$ and

$$B(X)=\prod_{1\leq r\leq n, r\neq i,j} (X+b_r-a_r) \tag{2}$$

is symmetrical in $$i$$ and $$j$$. It follows that

$$g(i)+g(j)=\frac{1}{d_{ij}}\Bigg(\frac{A(a_i-b_i)}{B(a_i-b_i)}- \frac{A(a_j-b_j)}{B(a_j-b_j)}\Bigg)$$ and $$g(i)+g(j)= \frac{\frac{A(a_i-b_i)B(a_j-b_j)-A(a_j-b_j)B(a_i-b_i)}{(a_i-b_i)-(a_j-b_j)}}{B(a_i-b_i)B(a_j-b_j)} \tag{3}$$

But the numerator in (3) above is in fact a polynomial. So, $$d_{ij}$$ does not appear in the denominator of the reduced fraction for $$g(i)+g(j)$$. Since $$d_{ij}$$ does not appear in the denominator of $$g(m)$$ when $$m\neq i,j$$, it follows that $$d_{i,j}$$ does not appear in the denominator of the reduced fraction for $$S$$.

Since this holds for any $$i,j$$,it follows that $$S$$ is a polynomial. Since each $$g(m)$$ has degree at most $$1$$ in each of its variables (as a rational fraction), $$S$$ also has degree at most $$1$$ in each of its variables.

So $$S$$ can be written

$$S=C(a_1,a_2,\ldots,a_l)+\sum_{m=1}^{l} S_m(a_1,a_2,\ldots,a_l) b_m \tag{4}$$

By taking all the $$b_k$$'s equal to $$0$$, we see that $$C=0$$. If we take all the $$b_k$$'s equal to $$0$$ except $$b_1$$, we see that $$g(1)=a_1b_1$$ and $$g(m)=0$$ for $$m>1$$, whence $$S_1=a_1$$.

More generally, if all the $$b_k$$'s are equal to $$0$$ except $$b_j$$ for $$j\gt 1$$, then $$g(m)=0$$ except when $$m=j$$ or $$j-1$$, with $$g(j-1)=\frac{b_ja_{j-1}(-a_{j-1}+a_j)}{a_{j-1}-a_j+b_j}$$ and a similar formula for $$g(j)$$, so that $$S_j=\frac{g(j-1)+g(j)}{b_j}=a_{j}-a_{j-1}$$.

We prove the result using induction. It is already verified that the result holds for $$l=2$$. So assume that the following holds: $$\begin{equation} \sum_{m=1}^l g(m,l)=\sum_{t=1}^l (a_t-a_{t+1})b_t. \end{equation}$$ We have \begin{align} \sum_{m=1}^{l+1} g(m,l+1) &= -\sum_{m=1}^{l+1}\frac{\prod_{r=1}^{l}a_m-a_r+b_{r+1}-b_m}{\prod_{r=1,r\neq m}^{l+1}a_m-a_r+b_{r}-b_m}(a_m+b_1-b_m)(a_m-a_{l+1}-b_m)\\ &= \sum_{m=1}^{l} g(m,l)\frac{(a_m-a_l+b_{l+1}-b_m)(a_m-a_{l+1}-b_m)}{(a_m-a_{l+1}+b_{l+1}-b_m)(a_m-a_l-b_m)}+g(l+1,l+1)\\ &= \sum_{m=1}^{l} g(m,l)\left(1-\frac{(a_{l+1}-a_l)b_{l+1}}{(a_m-a_{l+1}+b_{l+1}-b_m)(a_m-a_l-b_m)}\right)+g(l+1,l+1)\\ &= \sum_{t=1}^l(a_t-a_{t-1})b_{t}+(a_{l+1}-a_l)b_{l+1}P, \end{align} where $$\begin{equation} P=-\sum_{m=1}^{l} \frac{g(m,l)}{(a_m-a_{l+1}+b_{l+1}-b_m)(a_m-a_l-b_m)}+\frac{g(l+1,l+1)}{(a_{l+1}-a_l)b_{l+1}}. \end{equation}$$ Now it suffices to prove that $$P=1$$.
Next, we simplify $$g(l+1,l+1)$$: \begin{align} \frac{g(l+1,l+1)}{(a_{l+1}-a_l)b_{l+1}}& = -\frac{1}{(a_{l+1}-a_l)b_{l+1}}\frac{\prod_{r=1}^{l}a_{l+1}-a_r+b_{r+1}-b_{l+1}}{\prod_{r=1}^{l}a_{l+1}-a_r+b_{r}-b_{l+1}}(a_{l+1}+b_1-b_{l+1})b_{l+1}\\& = -(a_{l+1}+b_1-b_{l+1})f(x)|_{x=a_{l+1}-b_{l+1}} ,, \end{align} where $$f(x)=-\frac{\prod_{r=1}^{l-1}(x-a_r+b_{r+1})}{\prod_{r=1}^l x-a_r+b_{r}}$$. Using partial fractions, $$\begin{equation} f(x)=\sum_{p=1}^l\frac{c_p}{x-a_p+b_{p}} = \frac{\sum_{p=1}^l \left(c_p \prod_{r=1,r\neq p}^lx-a_r+b_{r} \right)}{\prod_{r=1}^l x-a_r+b_{r}}. \end{equation}$$ Thus, we get \begin{align} -\prod_{r=1}^{l-1}(x-a_r+b_{r+1})= \sum_{p=1}^l c_p \prod_{r=1,r\neq p}^l (x-a_r+b_{r})\hspace{5cm}(1) \end{align} Substituting $$x=a_m-b_m$$ gives the following: $$\begin{equation} c_m = -\frac{\prod_{r=1}^{l-1}a_m-a_r+b_{r+1}-b_m}{ \prod_{r=1,r\neq m}^l a_m-a_r+b_{r}-b_m} = g(m,l)\frac{1}{(a_m+b_1-b_m)(a_m-a_l-b_m)}. \end{equation}$$ Substituting this back in the expression for $$P$$, we get \begin{align} P &= -\sum_{m=1}^{l} \frac{g(m,l)}{(a_m-a_{l+1}+b_{l+1}-b_m)(a_m-a_l-b_m)}\\ &\hspace{1cm}-(a_{l+1}+b_1-b_{l+1})\sum_{m=1}^l g(m,l)\frac{1}{(a_{l+1}-a_m+b_{m}-b_{l+1})(a_m+b_1-b_m)(a_m-a_l-b_m)}\\ &= -\sum_{m=1}^{l} \frac{g(m,l)}{(a_m-a_{l+1}+b_{l+1}-b_m)(a_m-a_l-b_m)}\left(1-\frac{a_{l+1}+b_1-b_{l+1}}{a_m+b_1-b_m}\right)\\ &= -\sum_{m=1}^{l} g(m,l)\frac{1}{(a_m-a_l-b_m)(a_m+b_1-b_m)}= -\sum_{m=1}^{l} c_m = 1. \end{align} Here, the last equation follows because $$\sum_{m=1}^{l} c_m$$ is the coefficient of $$x$$ in (1).