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+5
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avatar+2353 

Hi there :D!

 

Can someone please help me figure out this equation?

 

Thanks in advance!

difficulty advanced
reinout-g  Jan 9, 2015

Best Answer 

 #2
avatar+20633 
+10

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

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

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

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

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

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

heureka  Jan 9, 2015
 #1
avatar+11852 
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?)

rosala  Jan 9, 2015
 #2
avatar+20633 
+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

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