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Hi there :D!

Jan 9, 2015

#2
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$$\small{\text{ I.  \sum \limits_{m=0}^{n-1}b \left( \prod \limits_{j=m+1}^{n-1}a \right)=b\sum \limits_{m=0}^{n-1} \left( \prod \limits_{j=m+1}^{n-1}a \right) \quad  because:  \quad b*(p_1)+b*(p_2)+b*(p_3)+... = b*(p_1+p_2+p_3+...) }}\\\\$$$$$\\\small{\text{ II. \prod \limits_{j=m+1}^{n-1}(a) \right)=a^{(n-1)-(m+1)+1} \quad because: \ \prod \limits_{from}^{to}(a) \right)= a^{\text{to}-\text{from}+1} \quad example: \ \prod \limits_{4}^{5}(a) \right)= a*a=a^{5-4+1}=a^2 }}\\ \small{\text{ \prod \limits_{j=m+1}^{n-1}(a) \right)=a^{(n-1)-m} }}\\\\$$$

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

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

$$\small{\text{ Result:  \sum \limits_{m=0}^{n-1}b \left( \prod \limits_{j=m+1}^{n-1}a \right)=b\sum \limits_{m=0}^{n-1} \left( \prod \limits_{j=m+1}^{n-1}a \right) =b*\dfrac{1-a^n}{ 1-a }  }}\\\\$$$. Jan 9, 2015 2+0 Answers #1 +11912 0 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?) Jan 9, 2015 #2 +26319 +10 Best Answer $$\small{\text{ I. \sum \limits_{m=0}^{n-1}b \left( \prod \limits_{j=m+1}^{n-1}a \right)=b\sum \limits_{m=0}^{n-1} \left( \prod \limits_{j=m+1}^{n-1}a \right) \quad because: \quad b*(p_1)+b*(p_2)+b*(p_3)+... = b*(p_1+p_2+p_3+...) }}\\\\$$$

$$\\\small{\text{ II. \prod \limits_{j=m+1}^{n-1}(a) \right)=a^{(n-1)-(m+1)+1} \quad because: \ \prod \limits_{from}^{to}(a) \right)= a^{\text{to}-\text{from}+1} \quad  example: \ \prod \limits_{4}^{5}(a) \right)= a*a=a^{5-4+1}=a^2 }}\\ \small{\text{ \prod \limits_{j=m+1}^{n-1}(a) \right)=a^{(n-1)-m} }}\\\\$$$$$\small{\text{ III. b \sum\limits_{m=0}^{n-1}a^{(n-1-m)} \right) =b \sum\limits_{m=0}^{n-1}a^{(m)} \quad because: a^{n-1}+a^{n-2}+\dots+a^1+a^0 = a^0+a^1+\dots + a^{n-2}+a^{n-1} }}\\\\$$$

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

heureka Jan 9, 2015