equation

$$\small{\text{
I.
$
\sum \limits_{m=0}^{n-1}b \left( \prod \limits_{j=m+1}^{n-1}a \right)=b\sum \limits_{m=0}^{n-1} \left( \prod \limits_{j=m+1}^{n-1}a \right)
\quad $
because: $ \quad b*(p_1)+b*(p_2)+b*(p_3)+... = b*(p_1+p_2+p_3+...)$
}}$\\\\$$$

$$\\\small{\text{
II.
$\prod \limits_{j=m+1}^{n-1}(a) \right)=a^{(n-1)-(m+1)+1} $\quad
because: $\ \prod \limits_{from}^{to}(a) \right)= a^{\text{to}-\text{from}+1} \quad $ example: $\ \prod \limits_{4}^{5}(a) \right)= a*a=a^{5-4+1}=a^2$
}}$\\$
\small{\text{
$\prod \limits_{j=m+1}^{n-1}(a) \right)=a^{(n-1)-m}$
}}$\\\\$$$

$$\small{\text{
III.
$
b \sum\limits_{m=0}^{n-1}a^{(n-1-m)} \right)
=b \sum\limits_{m=0}^{n-1}a^{(m)} \quad $ because: $a^{n-1}+a^{n-2}+\dots+a^1+a^0 = a^0+a^1+\dots + a^{n-2}+a^{n-1}$
}}$\\\\$$$

$$\small{\text{
IV. \ m = l:
$ \quad b \sum\limits_{m=0}^{n-1}a^{(m)} =
b \sum\limits_{l=0}^{n-1}a^{(l)} $
}}$\\\\$$$

$$\\\small{\text{
V.
$ \sum\limits_{l=0}^{n-1}a^{(l)} = a^0+a^1+a^2+\dots + a^{n-2}+a^{n-1} \quad $ this is a geometric series
}}$\\\\$
\small{\text{
The sum $s$ is:
$
a^0+a^1+a^2+\dots + a^{n-2}+a^{n-1}
$
}}$\\$
\small{\text{
$a*s$ is:
$
a^1+a^2+\dots + a^{n-1}+a^n
$
}}$\\$
\small{\text{
$s-a*s$ is:
$
a^0-a^n=1-a^n
$
}}$\\$
\small{\text{
$s(1-a)=1-a^n
$
}}$\\$
\small{\text{
$s=\frac{1-a^n}{ 1-a }$
}}$\\$
\small{\text{
$\sum\limits_{l=0}^{n-1}a^{(l)} = \dfrac{1-a^n}{ 1-a }
$
}}$$

$$\small{\text{
Result:
$
\sum \limits_{m=0}^{n-1}b \left( \prod \limits_{j=m+1}^{n-1}a \right)=b\sum \limits_{m=0}^{n-1} \left( \prod \limits_{j=m+1}^{n-1}a \right)
=b*\dfrac{1-a^n}{ 1-a }
$
}}$\\\\$$$

Oh at least we got a sight of you becoz of the question you asked!

as you might be aware this question is above my ahead so I can not do much . I hope you get an answer soon so Good Luck reinout! Btw when is your restaurant going to open?( lunch boxes?)