Mixing
Formula for series (or sequence?) of $n$ values where last in series is $x times n_1$ and total is $y$.
I am a programmer, not a mathematician and I'm trying to solve a problem for scoring in my game. Please excuse my weak maths and lack of mathematical syntax, and sorry if this question has been asked before.
Let's say we have a game level which consists of 4 stages. As the players moves from stage to stage, the required score for each stage increases. Let's say the overall score for the level is 1000 and I want the last stage to require 4 times more score.
Let these be the variables:
$y$ is level total required score.
$n_1, n_2, n_3, n_4$ are 4 stages in level.
$x$ is the score requirement multiplier for the last stage ($d$).
So to represent as an equation:
$n_1 + n_2 + n_3 + n_4 = y$ where $n_4 = x times n_1$
therefore based on the values in my example the equation would be:
$n_1 + n_2 + n_3 + 4n_1 = 1000$
How do I solve for $n_1$ , $n_2$ and $n_3$ such that there is a progression in the score requirement from stage to stage?
For some reason this formula works, substituting for the variables:
$n_1 + frac {4n_1}{3} + frac {4n_1}{2} + 4n_1 = 1000$
I able to solve for $n_1$ and I get $n_1 = 120$.
This answer checks out:
$120 + frac {4 times 120}{3} + frac {4 times 120}{2} + (4 times 120) = 1000$
$120 + 160 + 240 + 480 = 1000$
But I don't understand how that works. Where do the denominators for the fractions $frac 43$ and $frac 42$ come from?
Anyone that could shed some light on how to solve this, your time would be appreciated :)
sequences-and-series
edited Nov 22 '18 at 7:22
mlc
4,87931332
asked Nov 21 '18 at 9:09
jamo
162
add a comment |
I am a programmer, not a mathematician and I'm trying to solve a problem for scoring in my game. Please excuse my weak maths and lack of mathematical syntax, and sorry if this question has been asked before.
Let's say we have a game level which consists of 4 stages. As the players moves from stage to stage, the required score for each stage increases. Let's say the overall score for the level is 1000 and I want the last stage to require 4 times more score.
Let these be the variables:
$y$ is level total required score.
$n_1, n_2, n_3, n_4$ are 4 stages in level.
$x$ is the score requirement multiplier for the last stage ($d$).
So to represent as an equation:
$n_1 + n_2 + n_3 + n_4 = y$ where $n_4 = x times n_1$
therefore based on the values in my example the equation would be:
$n_1 + n_2 + n_3 + 4n_1 = 1000$
How do I solve for $n_1$ , $n_2$ and $n_3$ such that there is a progression in the score requirement from stage to stage?
For some reason this formula works, substituting for the variables:
$n_1 + frac {4n_1}{3} + frac {4n_1}{2} + 4n_1 = 1000$
I able to solve for $n_1$ and I get $n_1 = 120$.
This answer checks out:
$120 + frac {4 times 120}{3} + frac {4 times 120}{2} + (4 times 120) = 1000$
$120 + 160 + 240 + 480 = 1000$
But I don't understand how that works. Where do the denominators for the fractions $frac 43$ and $frac 42$ come from?
Anyone that could shed some light on how to solve this, your time would be appreciated :)
sequences-and-series
edited Nov 22 '18 at 7:22
mlc
4,87931332
asked Nov 21 '18 at 9:09
jamo
162
3
It looks like a geometric sum. You're starting with 120. Then you add 40, and then $2times 40$ and then $2^2 times 40 ...$ So you're basically calculating $$ 120 + 40 sum_{k=0}^n 2^k $$
– Matti P.
Nov 21 '18 at 9:18
Thanks @MattiP. I wish I could read that formula... :) I think I understand the gist of it.
– jamo
Nov 21 '18 at 9:36
add a comment |
I am a programmer, not a mathematician and I'm trying to solve a problem for scoring in my game. Please excuse my weak maths and lack of mathematical syntax, and sorry if this question has been asked before.
Let's say we have a game level which consists of 4 stages. As the players moves from stage to stage, the required score for each stage increases. Let's say the overall score for the level is 1000 and I want the last stage to require 4 times more score.
Let these be the variables:
$y$ is level total required score.
$n_1, n_2, n_3, n_4$ are 4 stages in level.
$x$ is the score requirement multiplier for the last stage ($d$).
So to represent as an equation:
$n_1 + n_2 + n_3 + n_4 = y$ where $n_4 = x times n_1$
therefore based on the values in my example the equation would be:
$n_1 + n_2 + n_3 + 4n_1 = 1000$
How do I solve for $n_1$ , $n_2$ and $n_3$ such that there is a progression in the score requirement from stage to stage?
For some reason this formula works, substituting for the variables:
$n_1 + frac {4n_1}{3} + frac {4n_1}{2} + 4n_1 = 1000$
I able to solve for $n_1$ and I get $n_1 = 120$.
This answer checks out:
$120 + frac {4 times 120}{3} + frac {4 times 120}{2} + (4 times 120) = 1000$
$120 + 160 + 240 + 480 = 1000$
But I don't understand how that works. Where do the denominators for the fractions $frac 43$ and $frac 42$ come from?
Anyone that could shed some light on how to solve this, your time would be appreciated :)
sequences-and-series
edited Nov 22 '18 at 7:22
mlc
4,87931332
asked Nov 21 '18 at 9:09
jamo
162
I am a programmer, not a mathematician and I'm trying to solve a problem for scoring in my game. Please excuse my weak maths and lack of mathematical syntax, and sorry if this question has been asked before.
Let's say we have a game level which consists of 4 stages. As the players moves from stage to stage, the required score for each stage increases. Let's say the overall score for the level is 1000 and I want the last stage to require 4 times more score.
Let these be the variables:
$y$ is level total required score.
$n_1, n_2, n_3, n_4$ are 4 stages in level.
$x$ is the score requirement multiplier for the last stage ($d$).
So to represent as an equation:
$n_1 + n_2 + n_3 + n_4 = y$ where $n_4 = x times n_1$
therefore based on the values in my example the equation would be:
$n_1 + n_2 + n_3 + 4n_1 = 1000$
How do I solve for $n_1$ , $n_2$ and $n_3$ such that there is a progression in the score requirement from stage to stage?
For some reason this formula works, substituting for the variables:
$n_1 + frac {4n_1}{3} + frac {4n_1}{2} + 4n_1 = 1000$
I able to solve for $n_1$ and I get $n_1 = 120$.
This answer checks out:
$120 + frac {4 times 120}{3} + frac {4 times 120}{2} + (4 times 120) = 1000$
$120 + 160 + 240 + 480 = 1000$
But I don't understand how that works. Where do the denominators for the fractions $frac 43$ and $frac 42$ come from?
Anyone that could shed some light on how to solve this, your time would be appreciated :)
sequences-and-series
sequences-and-series
edited Nov 22 '18 at 7:22
mlc
4,87931332
asked Nov 21 '18 at 9:09
jamo
162
edited Nov 22 '18 at 7:22
mlc
4,87931332
asked Nov 21 '18 at 9:09
jamo
162
edited Nov 22 '18 at 7:22
mlc
4,87931332
edited Nov 22 '18 at 7:22
mlc
4,87931332
edited Nov 22 '18 at 7:22
mlc
4,87931332
4,87931332
asked Nov 21 '18 at 9:09
jamo
162
asked Nov 21 '18 at 9:09
jamo
162
asked Nov 21 '18 at 9:09
jamo
162
162
3
It looks like a geometric sum. You're starting with 120. Then you add 40, and then $2times 40$ and then $2^2 times 40 ...$ So you're basically calculating $$ 120 + 40 sum_{k=0}^n 2^k $$
– Matti P.
Nov 21 '18 at 9:18
Thanks @MattiP. I wish I could read that formula... :) I think I understand the gist of it.
– jamo
Nov 21 '18 at 9:36
add a comment |
3
It looks like a geometric sum. You're starting with 120. Then you add 40, and then $2times 40$ and then $2^2 times 40 ...$ So you're basically calculating $$ 120 + 40 sum_{k=0}^n 2^k $$
– Matti P.
Nov 21 '18 at 9:18
Thanks @MattiP. I wish I could read that formula... :) I think I understand the gist of it.
– jamo
Nov 21 '18 at 9:36
3
3
It looks like a geometric sum. You're starting with 120. Then you add 40, and then $2times 40$ and then $2^2 times 40 ...$ So you're basically calculating $$ 120 + 40 sum_{k=0}^n 2^k $$
– Matti P.
Nov 21 '18 at 9:18
It looks like a geometric sum. You're starting with 120. Then you add 40, and then $2times 40$ and then $2^2 times 40 ...$ So you're basically calculating $$ 120 + 40 sum_{k=0}^n 2^k $$
– Matti P.
Nov 21 '18 at 9:18
Thanks @MattiP. I wish I could read that formula... :) I think I understand the gist of it.
– jamo
Nov 21 '18 at 9:36
Thanks @MattiP. I wish I could read that formula... :) I think I understand the gist of it.
– jamo
Nov 21 '18 at 9:36
add a comment |
1 Answer
1
active
oldest
votes
This question is a little difficult to answer mathematically, since $a+b+c+xa=1000$ has many possible solutions. So let's refine it a little bit. Let's add a requirement that each stage costs $r$ times as much as the previous stage. We can then rewrite the equation as
$$a+ar+ar^2+ar^3=1000.$$
At this point, you can just pick your favorite value of $r$ and solve for $a$. If you want each stage to cost $2$ times as much as the previous stage then
$$a+2a+4a+8a=1000$$
$$15a=1000$$
$$a=frac{200}3 approx 67.$$
This gives the value of the first stage, and then $2a, 4a,$ and $8a$ are the values of the stages afterward.
answered Nov 21 '18 at 9:18
DreamConspiracy
9001216
That makes much more sense - to define the progression amount, not the final multiple. Which means my starting point was wrong, making this really hard to solve. Thanks so much for your answer :)
– jamo
Nov 21 '18 at 9:28
add a comment |
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1 Answer
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1 Answer
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votes
This question is a little difficult to answer mathematically, since $a+b+c+xa=1000$ has many possible solutions. So let's refine it a little bit. Let's add a requirement that each stage costs $r$ times as much as the previous stage. We can then rewrite the equation as
$$a+ar+ar^2+ar^3=1000.$$
At this point, you can just pick your favorite value of $r$ and solve for $a$. If you want each stage to cost $2$ times as much as the previous stage then
$$a+2a+4a+8a=1000$$
$$15a=1000$$
$$a=frac{200}3 approx 67.$$
This gives the value of the first stage, and then $2a, 4a,$ and $8a$ are the values of the stages afterward.
answered Nov 21 '18 at 9:18
DreamConspiracy
9001216
That makes much more sense - to define the progression amount, not the final multiple. Which means my starting point was wrong, making this really hard to solve. Thanks so much for your answer :)
– jamo
Nov 21 '18 at 9:28
add a comment |
This question is a little difficult to answer mathematically, since $a+b+c+xa=1000$ has many possible solutions. So let's refine it a little bit. Let's add a requirement that each stage costs $r$ times as much as the previous stage. We can then rewrite the equation as
$$a+ar+ar^2+ar^3=1000.$$
At this point, you can just pick your favorite value of $r$ and solve for $a$. If you want each stage to cost $2$ times as much as the previous stage then
$$a+2a+4a+8a=1000$$
$$15a=1000$$
$$a=frac{200}3 approx 67.$$
This gives the value of the first stage, and then $2a, 4a,$ and $8a$ are the values of the stages afterward.
answered Nov 21 '18 at 9:18
DreamConspiracy
9001216
That makes much more sense - to define the progression amount, not the final multiple. Which means my starting point was wrong, making this really hard to solve. Thanks so much for your answer :)
– jamo
Nov 21 '18 at 9:28
add a comment |
This question is a little difficult to answer mathematically, since $a+b+c+xa=1000$ has many possible solutions. So let's refine it a little bit. Let's add a requirement that each stage costs $r$ times as much as the previous stage. We can then rewrite the equation as
$$a+ar+ar^2+ar^3=1000.$$
At this point, you can just pick your favorite value of $r$ and solve for $a$. If you want each stage to cost $2$ times as much as the previous stage then
$$a+2a+4a+8a=1000$$
$$15a=1000$$
$$a=frac{200}3 approx 67.$$
This gives the value of the first stage, and then $2a, 4a,$ and $8a$ are the values of the stages afterward.
answered Nov 21 '18 at 9:18
DreamConspiracy
9001216
This question is a little difficult to answer mathematically, since $a+b+c+xa=1000$ has many possible solutions. So let's refine it a little bit. Let's add a requirement that each stage costs $r$ times as much as the previous stage. We can then rewrite the equation as
$$a+ar+ar^2+ar^3=1000.$$
At this point, you can just pick your favorite value of $r$ and solve for $a$. If you want each stage to cost $2$ times as much as the previous stage then
$$a+2a+4a+8a=1000$$
$$15a=1000$$
$$a=frac{200}3 approx 67.$$
This gives the value of the first stage, and then $2a, 4a,$ and $8a$ are the values of the stages afterward.
answered Nov 21 '18 at 9:18
DreamConspiracy
9001216
answered Nov 21 '18 at 9:18
DreamConspiracy
9001216
answered Nov 21 '18 at 9:18
DreamConspiracy
9001216
answered Nov 21 '18 at 9:18
DreamConspiracy
9001216
9001216
That makes much more sense - to define the progression amount, not the final multiple. Which means my starting point was wrong, making this really hard to solve. Thanks so much for your answer :)
– jamo
Nov 21 '18 at 9:28
add a comment |
That makes much more sense - to define the progression amount, not the final multiple. Which means my starting point was wrong, making this really hard to solve. Thanks so much for your answer :)
– jamo
Nov 21 '18 at 9:28
That makes much more sense - to define the progression amount, not the final multiple. Which means my starting point was wrong, making this really hard to solve. Thanks so much for your answer :)
– jamo
Nov 21 '18 at 9:28
That makes much more sense - to define the progression amount, not the final multiple. Which means my starting point was wrong, making this really hard to solve. Thanks so much for your answer :)
– jamo
Nov 21 '18 at 9:28
add a comment |
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3
It looks like a geometric sum. You're starting with 120. Then you add 40, and then $2times 40$ and then $2^2 times 40 ...$ So you're basically calculating $$ 120 + 40 sum_{k=0}^n 2^k $$
– Matti P.
Nov 21 '18 at 9:18
Thanks @MattiP. I wish I could read that formula... :) I think I understand the gist of it.
– jamo
Nov 21 '18 at 9:36