Explain how you would find the weight of each stacked box if you knew the weight of the bottom box. Find the weight of each box in a stack of 4 boxes
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eddy2017
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Okay. That is good. I take a read of that. I'll take a break now, Doc. I need it. Thank you. I'll go back to the problem and study it from that different perspective of a derived formulaProgression is a property - we have progression in Mathematics, in Music, in Life ...... all finite. It has nothing to do with finiteness or infiniteness. You may want to study:
Progression - Wikipedia
en.wikipedia.org
First, Eddy, I am not a mathematician at all.
My academic training in college was in European languages and history. My first real job was as a computer programmer; coding is a quasi-mathematical language. My training in graduate school was in economics, which involves some math. So I picked up basic mathematics along the way, particularly at my prep school, which focused almost exclusively on languages and math.
Now I do not know anything about the site you referenced, but what they are showing is not the standard presentation of a geometric series. The standard presentation looks like this.
[math]w + rw + r^2w + \ … \ r^nw = w + w * \sum_{k=1}^n r^k = w * \sum_{k=0}^n r^k.[/math]
Notice this has n + 1 terms and does not start with r. We can of course have a special case where w = r. It looks as though that special case is what is going on in the (1/2) + (1/4) + (1/8) example.
I prefer the standard presentation because it leads to this formula.
[math]r \ne 1, n \ge 2, \text { and } t = w * \sum_{k=0} r^k.[/math]
[math]\therefore rt = w * \sum_{k=0}^n r^{(k+1)} = \left (w * \sum_{k=1}^{(n-1)} r^k \right ) + wr^{(n+1)}.[/math]
And we can restate t as
[math]t = w * r^0 + \left ( w * \sum_{k=1}^n \right ) = w + \left ( w * \sum_{k=1}^n r^k \right ).[/math]
[math]\therefore t - rt = w + \left ( w * \sum_{k=1}^n r^k \right ) - \left \{ \left (w * \sum_{k=1}^n r^k \right ) + wr^{(n+1)} \right \} = w- wr^{(n+1)} = w(1 - r^{(n+1)}.) [/math]
[math]\text {And } t - rt = t(1 - r). [math]\therefore t = w * \dfrac{1 - r^{(n+1)}}{1 - r} \text { because } r \ne 1..[/math] ....[edited............was missing the 'w']
Once you understand HOW the formula is derived, you will understand WHEN to use it properly.
My academic training in college was in European languages and history. My first real job was as a computer programmer; coding is a quasi-mathematical language. My training in graduate school was in economics, which involves some math. So I picked up basic mathematics along the way, particularly at my prep school, which focused almost exclusively on languages and math.
Now I do not know anything about the site you referenced, but what they are showing is not the standard presentation of a geometric series. The standard presentation looks like this.
[math]w + rw + r^2w + \ … \ r^nw = w + w * \sum_{k=1}^n r^k = w * \sum_{k=0}^n r^k.[/math]
Notice this has n + 1 terms and does not start with r. We can of course have a special case where w = r. It looks as though that special case is what is going on in the (1/2) + (1/4) + (1/8) example.
I prefer the standard presentation because it leads to this formula.
[math]r \ne 1, n \ge 2, \text { and } t = w * \sum_{k=0} r^k.[/math]
[math]\therefore rt = w * \sum_{k=0}^n r^{(k+1)} = \left (w * \sum_{k=1}^{(n-1)} r^k \right ) + wr^{(n+1)}.[/math]
And we can restate t as
[math]t = w * r^0 + \left ( w * \sum_{k=1}^n \right ) = w + \left ( w * \sum_{k=1}^n r^k \right ).[/math]
[math]\therefore t - rt = w + \left ( w * \sum_{k=1}^n r^k \right ) - \left \{ \left (w * \sum_{k=1}^n r^k \right ) + wr^{(n+1)} \right \} = w- wr^{(n+1)} = w(1 - r^{(n+1)}.) [/math]
[math]\text {And } t - rt = t(1 - r). [math]\therefore t = w * \dfrac{1 - r^{(n+1)}}{1 - r} \text { because } r \ne 1..[/math] ....[edited............was missing the 'w']
Once you understand HOW the formula is derived, you will understand WHEN to use it properly.
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