# Show that $\prod_{i=1}^n a_i- \prod_{j=1}^n b_i =$ $\sum_{t=1}^{n-1}(\prod_{i\leq t-1}a_i)(\prod_{j\geq t+1} b_j)(a_t-b_t)$

Pardon the cryptic notation and possibly trivial question. I believe the following holds.

Define $$X_t=(\prod_{i\leq t-1}a_i)(\prod_{j\geq t+1} b_j)(a_t-b_t).$$ Show that $$\prod_{i=1}^na_i-\prod_{j=1}^nb_i=\sum_{t=1}^{n-1}X_t.$$

Is there a quick and elegant way of seeing this? It's straightforward by induction and I think as well by multilinearity of determinants and possibly even by inclusion-exclusion. Moreover, it should be also possible from expanding $(a_1-b_1)(a_2-b_2)...(a_{n-1}-b_{n-1})(a_n+(-1)^nb_n)$.

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Assume first that $b_t=1$ for every $t$ and introduce $A_t=\prod\limits_{i=1}^ta_i$, in particular $A_0=1$ and $A_t=a_tA_{t-1}$ for every $1\leqslant t\leqslant n$. Thus, the formula $X_t=A_{t-1}\cdot(a_t-1)=A_t-A_{t-1}$ yields the telescoping series $\sum\limits_{t=1}^nX_t=A_n-A_0=A_n-1$. This is the expression to prove, except that the sum should extend up to $t=n$.
In the general case, introduce $A_t=\prod\limits_{i=1}^ta_i\cdot\prod\limits_{j=t+1}^nb_j$, in particular $A_0=\prod\limits_{t=1}^nb_t$, $A_n=\prod\limits_{t=1}^na_t$ and $A_t=c_t\cdot A_{t-1}$ with $c_t=a_t/b_t$. Thus, the formula $X_t=A_{t-1}\cdot(c_t-1)=A_t-A_{t-1}$ yields the telescoping series $\sum\limits_{t=1}^nX_t=A_n-A_0=\prod\limits_{t=1}^na_t-\prod\limits_{t=1}^nb_t$. This is the expression to prove, except that the sum should extend up to $t=n$.