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

[eddy2017]

Okay , that explains it better. So those were derived formulas. Okay, thank you Dr Khan. Okay, I was fixed on finding the formula. That might have been it.

 

[eddy2017]

Progression 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

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 formula

 

[JeffM]

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.

$$w + rw + r^2w + \ … \ r^nw = w + w * \sum_{k=1}^n r^k = w * \sum_{k=0}^n r^k.$$
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.

$$r \ne 1, n \ge 2, \text { and } t = w * \sum_{k=0} r^k.$$
$$\therefore rt = w * \sum_{k=0}^n r^{(k+1)} = \left (w * \sum_{k=1}^{(n-1)} r^k \right ) + wr^{(n+1)}.$$
And we can restate t as

$$t = w * r^0 + \left ( w * \sum_{k=1}^n \right ) = w + \left ( w * \sum_{k=1}^n r^k \right ).$$
$$\therefore t - rt = w + \left ( w * \sum_{k=1}^n r^k \right ) - \left \{ \left (w * \sum_{k=1}^n r^k \right ) + wr^{(n+1)} \right \} = w- wr^{(n+1)} = w(1 - r^{(n+1)}.)$$
$$\text {And } t - rt = t(1 - r). [math]\therefore t = w * \dfrac{1 - r^{(n+1)}}{1 - r} \text { because } r \ne 1..$$ ....[edited............was missing the 'w']

Once you understand HOW the formula is derived, you will understand WHEN to use it properly.

 

[eddy2017]

Wow, Jeff, that was something else!. Thanks you so much for taking the time!!!.