Find the Total Cost for X Cards Bought | Help Writing an Equation

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Discussion Overview

The discussion revolves around formulating an equation to calculate the total cost of purchasing X cards, where the price of each card increases incrementally with each purchase. Participants explore how to derive a summation equation based on the individual costs of the cards, which are defined by a linear relationship.

Discussion Character

  • Mathematical reasoning
  • Technical explanation
  • Homework-related

Main Points Raised

  • One participant states that the first card costs 60 units, with each subsequent card increasing in price by 20 units, leading to a proposed individual card price equation of f(x) = 20X + 40.
  • Another participant suggests that the nth card can be expressed as 60 + (n-1) * 20, which simplifies to 40 + 20n.
  • A summation formula is proposed as the total cost = ∑(40 + 20n) from n = 1 to X.
  • There is a correction regarding a misinterpretation of the equation, clarifying that 60 + (n-1) * 20 equals 40 + 20n, and questioning the derivation of ∑(40 + 2n).
  • Participants express gratitude for assistance and acknowledge the need for clarification on the equations presented.

Areas of Agreement / Disagreement

Participants generally agree on the structure of the individual card pricing but have differing views on the summation expression and its derivation. The discussion remains unresolved regarding the exact formulation of the total cost equation.

Contextual Notes

There are unresolved questions about the derivation of certain equations and the assumptions underlying the pricing model. The discussion reflects varying levels of understanding of summation notation and its application in this context.

Colt1229
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Every time you buy a card the price goes up 20 units. The first card costs 60 units. How would you write an equation to find the TOTAL cost for X cards bought? I know how to get a price of an individual card by using the equation f(x)=20X+40. How would I write an equation that would add up the answers of f(x). I do not want to have to do it the long way.
The 1st card costs 60
2nd card costs 80
3rd costs 100
So the total cost for 3 cards is 240.
I do not know how to turn it into an equation though. If I wanted to go to 60 cards it would take forever.
 
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Colt1229 said:
Every time you buy a card the price goes up 20 units. The first card costs 60 units. How would you write an equation to find the TOTAL cost for X cards bought? I know how to get a price of an individual card by using the equation f(x)=20X+40. How would I write an equation that would add up the answers of f(x). I do not want to have to do it the long way.
The 1st card costs 60
2nd card costs 80
3rd costs 100
So the total cost for 3 cards is 240.
I do not know how to turn it into an equation though. If I wanted to go to 60 cards it would take forever.
The $1^{st}$ card costs 60
$2^{nd}$ card costs 80
the $n^{th}$ card costs 60 + (n-1) * 20 = 40 + 20n

so total cost = $\sum_{n= 1}^{X}(40+20n)$

you should be able to proceed
 
Last edited:
kaliprasad said:
The $1^{st}$ card costs 60
$2^{nd}$ card costs 80
the $n^{th}$ card costs 60 + (n-1) * 20 = 40 * 20n

so total cost = $\sum_{n= 1}^{X}(40+2n)$

you should be able to proceed

Not to sound rude, since i am asking for help. However, I do not think that 60+ (n-1)*20 is equal to 40*20n.

60+(n-1)*20 does however equal 40+20n. Also, I do not understand how you got \sum_{n=1}^{X}(40+2n). Well really I don't understand where the (40+2n) comes from. Thank you for helping me though!
 
Colt1229 said:
Not to sound rude, since i am asking for help. However, I do not think that 60+ (n-1)*20 is equal to 40*20n.

60+(n-1)*20 does however equal 40+20n. Also, I do not understand how you got \sum_{n=1}^{X}(40+2n). Well really I don't understand where the (40+2n) comes from. Thank you for helping me though!

Thanks. It was a mistake on my part 40 + 20n in place of 40* 2n.

now we have found the cost of $n^{th}$ item and we need to add the cost from 1st to X and hence the sum.
 
Thank you very much. You were a great help. I just needed a memory refresh because its been so long.
 

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