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?
 A: 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.
A: 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).
