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 said:
I was hoping for a clue. i have studied like 4 different equations now and none fits this
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What are those 4 equations?

Did you study their derivation ?

You won't need a clue if you can derive those equations - without looking at your reference!
 
I studied all related to geometric series and sequence. Just simply do not know which one applies here. I have tried several and all of them have be rejected
 
this one is the sum of the n terms of an arithmetic progression buut i have a finite series here not a progression
Sum of n terms of AP = n/2[2a + (n – 1)d]
 
Subhotosh Khan said:
What are those 4 equations?

Did you study their derivation ?

You won't need a clue if you can derive those equations - without looking at your reference!
Click to expand...
I have looked at my reference
i have written out all data given.
what else can I do?

WELL, TO ANSWER QUESTION 2
they are asking for the unknow wight of the bottom box now
x =(r^4-1)/(r-1)

this could be a possible way out but I just checked that the result when i replaced values is negative so it can not be
 
eddy2017 said:
One question. Should I simplify whatever is in brackets before i distribute?
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JeffM said:
Eddy

What is unknown? As the great Khan would say, what is the find?

ASSIGN A PRONUMERAL TO EACH.

b=weight of bottom box.b = \text {weight of bottom box}.
t=weight of stack.t = \text {weight of stack}.
WRITE IN MATH NOTATION ALL QUANTITATIVE RELATIONS GIVEN IN ENGLISH

[math]t = w + w * \dfrac{1}{2} + \left ( w * \dfrac{1}{2} \right ) * \dfrac{1}{2} + \left \{ \left (w * \dfrac{1}{2} \right ) * \dfrac{1}{2} \right \} * \dfrac{1}{2}\right \}.[/math]
Can you simplify that using the distributive property?

Do you know what that kind of series is called?

Do you know a formula for calculating the sum no matter how many terms are involved?
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I am going to go way back to my initial post. A major point in math is generalization. We learn techniques so we can apply them in many different situations.

So no, if I see a common factor in many terms I tend to simplify first by pulling the common factor out. Otherwise I may forget to include at some point and make an error.

t=w+w∗12+(w∗12)∗12+{(w∗12)∗12}∗12  ⟹  t = w + w * \dfrac{1}{2} + \left ( w * \dfrac{1}{2} \right ) * \dfrac{1}{2} + \left \{ \left (w * \dfrac{1}{2} \right ) * \dfrac{1}{2} \right \} * \dfrac{1}{2} \implies
t=w{1+12+12∗12+∗12∗12∗12)=w(1+0.5+0.25+0.125)=1.875w.t = w \left \{ 1 + \dfrac{1}{2} + \dfrac{1}{2} * \dfrac{1}{2} + * \dfrac{1}{2} * \dfrac{1}{2} * \dfrac{1}{2}\right ) = w ( 1 + 0.5 + 0.25 + 0.125) = 1.875w.
Now I can work with this problem no matter what weight the bottom box is.

It is even easier if I recognize this as a geometric series and know the formula. That will work with any common ratio and any number of terms.

w∗1−rn1−r=w{1−(12)4)÷12=2w(1−116)=15w8=1.875w.w * \dfrac{1 - r^n}{1 - r} = w \left \{1 - \left ( \dfrac{1}{2} \right )^4 \right) \div \dfrac{1}{2} = 2w \left ( 1 - \dfrac{1}{16} \right ) = \dfrac{15w}{8} = 1.875w.
 
JeffM said:
I am going to go way back to my initial post. A major point in math is generalization. We learn techniques so we can apply them in many different situations.

So no, if I see a common factor in many terms I tend to simplify first by pulling the common factor out. Otherwise I may forget to include at some point and make an error.

t=w+w∗12+(w∗12)∗12+{(w∗12)∗12}∗12  ⟹  t = w + w * \dfrac{1}{2} + \left ( w * \dfrac{1}{2} \right ) * \dfrac{1}{2} + \left \{ \left (w * \dfrac{1}{2} \right ) * \dfrac{1}{2} \right \} * \dfrac{1}{2} \implies
t=w{1+12+12∗12+∗12∗12∗12)=w(1+0.5+0.25+0.125)=1.875w.t = w \left \{ 1 + \dfrac{1}{2} + \dfrac{1}{2} * \dfrac{1}{2} + * \dfrac{1}{2} * \dfrac{1}{2} * \dfrac{1}{2}\right ) = w ( 1 + 0.5 + 0.25 + 0.125) = 1.875w.
Now I can work with this problem no matter what weight the bottom box is.

It is even easier if I recognize this as a geometric series and know the formula. That will work with any common ratio and any number of terms.

w∗1−rn1−r=w{1−(12)4)÷12=2w(1−116)=15w8=1.875w.w * \dfrac{1 - r^n}{1 - r} = w \left \{1 - \left ( \dfrac{1}{2} \right )^4 \right) \div \dfrac{1}{2} = 2w \left ( 1 - \dfrac{1}{16} \right ) = \dfrac{15w}{8} = 1.875w.
Click to expand...
According to this site, which I understand to be a good math site, this is a formula to find a geometric series. Is it correct or not?.
1640906343209.png
 
JeffM said:
I am going to go way back to my initial post. A major point in math is generalization. We learn techniques so we can apply them in many different situations.

So no, if I see a common factor in many terms I tend to simplify first by pulling the common factor out. Otherwise I may forget to include at some point and make an error.

t=w+w∗12+(w∗12)∗12+{(w∗12)∗12}∗12  ⟹  t = w + w * \dfrac{1}{2} + \left ( w * \dfrac{1}{2} \right ) * \dfrac{1}{2} + \left \{ \left (w * \dfrac{1}{2} \right ) * \dfrac{1}{2} \right \} * \dfrac{1}{2} \implies
t=w{1+12+12∗12+∗12∗12∗12)=w(1+0.5+0.25+0.125)=1.875w.t = w \left \{ 1 + \dfrac{1}{2} + \dfrac{1}{2} * \dfrac{1}{2} + * \dfrac{1}{2} * \dfrac{1}{2} * \dfrac{1}{2}\right ) = w ( 1 + 0.5 + 0.25 + 0.125) = 1.875w.
Now I can work with this problem no matter what weight the bottom box is.

It is even easier if I recognize this as a geometric series and know the formula. That will work with any common ratio and any number of terms.

w∗1−rn1−r=w{1−(12)4)÷12=2w(1−116)=15w8=1.875w.w * \dfrac{1 - r^n}{1 - r} = w \left \{1 - \left ( \dfrac{1}{2} \right )^4 \right) \div \dfrac{1}{2} = 2w \left ( 1 - \dfrac{1}{16} \right ) = \dfrac{15w}{8} = 1.875w.
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please,help me make head or tail of what was typed in here. why the brackets and why the braces?
what was the reason for such a statement.?. what is the fundamental premise? .Don't forget you are all high-end mathematicians talking a greenhorn in math. Would anyone care to explain the reason behind this?
Math is nothing without a reason. it is like whistling in a storm.
 
eddy2017 said:
this is a formula to find a geometric series
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I do not quite understand - what did you mean by "find a geometric series".

The expression derived is good for finding sum of a PARTICULAR geometric series - not a general sum.
 
Subhotosh Khan said:
I do not quite understand - what did you mean by "find a geometric series".

The expression derived is good for finding sum of a PARTICULAR geometric series - not a general sum.
Click to expand...
My bad. to find a term in a geometric series.
 
What is the formula that is a perfect fit to answer the second question of this exercise?
 
IT IS GOOD TO ASK PEOPLE TO SHOW WORK AND WATCH VIDEOS AND STUFF LIKE THAT. ANOTHER THING IS TO KEEP SOMEONE GUESSING FOR ANSWERS FOR ALMOST 24 HOURS. THAT IS CRAZY!.
 
eddy2017 said:
I have looked at my reference
i have written out all data given.
what else can I do?

WELL, TO ANSWER QUESTION 2
they are asking for the unknow wight of the bottom box now
x =(r^4-1)/(r-1)...........................................................................................................This is incorrect

this could be a possible way out but I just checked that the result when i replaced values is negative so it can not be
Click to expand...
 
hints tips, go here do this, try this, did not work, why dont you watch this and see if you get it, no, well, how about this go down this road. Holy toledo! Eddy, you need to use this formula, find yourselves the value. when people get stuck they need to get someone to get them unstuck

I have been at this for 12 hours believe you me!
 
I'LL TAKE A BREAK FOR THE NIGHT. HAVE A GOOD EVENING FELLOWS, AND THANK YOU!!!
 
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