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 $S_j=\frac{g(j-1)+g(j)}{b_j}=a_{j}-a_{j-1}$.
This proves your identity.
