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

[Colt1229]

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.

 

[kaliprasad]

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

 

[Colt1229]

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!

 

[kaliprasad]

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.

 

[Colt1229]

Thank you very much. You were a great help. I just needed a memory refresh because its been so long.