Question 307207: Today I've had a real though challenge of trying to solve this problem. At first my teacher couldn't solve it, so I asked some other math teachers and they couldn't either. Even in the teachers guide it did not have the answer, so I'm begging someone to please help me find the answer. I even asked some of the Geometry students who are two math levels above me and they couldn't solve it either! They are doing 10th or 11th grade math right now. So please, I beg some genius out there to help me solve this problem!
There are four plans so far.
On plan 1 the equation is r=2^*-1
The r in the equation stands for total rubas or just rubas.
In this case rubas are a coin from the ancient kingdom of Montarek uses.
The King and Queen both use this coin, because it is there money.
There is a chessboard with 64 squares on it.
One square 1, there is 1 ruba. On square 2, there are 2. On square 3, there are 4. On square 4, there are 8 rubas and so on. Keep going with this pattern until you get to 64 squares. You should then get the number 9.223372037 times 10 to the 18 power. That number is the number of rubas on the 64th square. So the equation would be r=2 to the x power minus 1. So you subtract an exponent. The x stands for an exponent. Every time there is another square, you double the amount of rubas on the previous square. So if there was 64 rubas on square 7, then there would be double that on the next square, square 8, which is 128. And you continue that pattern on until you get to 64 squares.
r=2^x-1
x=exponent
^=the power of
and then minus 1 exponent.
Then you have to figure out the total number of rubas combining all the number of rubas together on all 64 squares. What we got was around 1.85 times 10 to the 19th power. (We rounded the 1.85).
That whole serious was plan 1, an example of what I'm doing. I have already figured out the equation on how to get the anwswer to the problem. What I need to find out is a different plan, plan 2.
Plan 2 goes like this.
On square 1, there are a starting 20 rubas, and every time another square is added, 5 rubas are added to that square. There would be 25 rubas. Then on the 3rd square, there would be 30 and so forth. It goes up by 5 rubas everytime, but starts at 20 rubas on square 1. Remember that.
On square 64 there will be 335 rubas if you continue the pattern of going up by 5 every time. Then the next step gets tricky because you have to add all the amounts of rubas on each square, from 64 squares to get the total number of rubas. I added all of the rubas from each square together on my calculator to get the answer 11,360 which is correct. BUT now I need to figure out the equation of how to get to that number. Of how I added all the number of rubas on each square to get to 11,360.
There is also an equation my teacher gave us but doesn't work with this problem, only with the previous problem.
This is it:
s=Total
a=first amount
r=rate
n=number of squares
Equation:
s=a(1-1^x) / 1-r
r=5 because five is the rate.
Remember x is to a power like 1-1 to the 10th power for example.
What you are trying to solve is how to write an equation based on the plan of how to get 11,360 rubas alltogether. Please Help, I'm desperate!
I'm really not sure on how to do it because I tried it using the equation but it doesn't work. I've asked teachers and really smart students and can't remember how to. I even asked my own teacher and he can't solve it, so he's also been trying to solve it.
If you have any other question beyond all of this, PLEASE comment done below. And if you are able to solve this, you are a amazing genius!
Thanks for taking your time to read all of this! :)
Found 2 solutions by jim_thompson5910, scott8148:
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website! So your goal is to get 11,360 rubas total right? I'm not sure why you've made it this complicated, but why not just use an arithmetic sequence as you did in plan 2?
One arithmetic sequence that generates 20, 25, 30, ..., 330, 335 is  where 'n' starts at  . Notice that if  , then  which is the 3rd term.
Recall that the sum of the first n terms of an arithmetic sequence is  , where 'a' is the first term,  is the nth term and 'n' is the number of terms you want to sum up. In this case, there are 64 terms meaning that  , the first term is  and the last term is  .
Plug these values in to get 
So the sum of the terms 20, 25, 30, ..., 330, 335 is 11,360
Answer by scott8148(6628)  (Show Source):
You can put this solution on YOUR website! these are two different types of series, so the solution techniques are different
plan 1 is a GEOMETRIC series, any term equals the previous term MULTIPLIED by some constant (in this case, 2)
plan 2 is an ARITHMETIC series, any term equals the previous term with some constant ADDED (in this case, 5)
the sum of the arithmetic series is straightforward, if you think about it
the first term is some number,"a"
___ to find any term, "n", the equation is ___ a + (n-1)d
___ where "n" is the number of the term and "d" is the difference (added constant)
now look at plan 2 ___ a = 20 ___ d = 5
___ 1st term is 20 and 64th term is [20 + (63 * 5)] or 335 ___ the sum of these terms is 355
___ 2nd term is 25 and 63rd term is 330 ___ the sum of these terms is 355
see the pattern
for a series of N terms ___ take the sum of the pairs of terms (first and last)
___ divide by two to get an "average" term value
___ then multiply by the number of terms to get the sum of the series
___ s = ([a + (a + [(N-1)d])] / 2) * N = aN + ([(N^2-N)d] / 2)
for plan 2 ___ s = 20(64) + ([(4096 - 64)5] / 2) = 1280 + 10080 = 11360
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