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?)