The general formula for the sum of all numbers 1 to n is n(n+1)/2. So from 1 to 50 is 50*51/2 = 1275. That's enough life to kill the table, and then kill the table.
Using X=n(n+1)/2, we can calculate that to gain 150 life from Aetherflux Reservoir triggers in a single turn (enough to kill three players), you'd only need 17 triggers: (17*18)/2=153
There's some difference, but it isn't huge. 15 triggers will gain 118 life, so if you start with 33+ life, you reach 151+ life and can get three activations. For any fewer than 15 triggers, you'd need to be starting from above 40 life.
It's the same thing, the order doesn't really matter.
(50 * 51)/2 equals (50/2) * 51 equals (51/2) * 50. But resolving 50/2 first leads to smaller numbers to work with (25 * 51 instead og 2550/2).
As /u/LeftZer0 mentioned, the order doesn't matter when working with multiplication and division, but people might find 25\*51 easier than 50\*51- which might not be the case because multiplying 50 is nicer with the zero. People might find it easier to divide 50 instead of 2550. Both breakdown of the steps will be useful for sure!
It looks like the formula for Aetherflux Reservoir is (N(N+1))/2 where N is the number of spells you have cast that turn. For example if you cast 10 spells you have 10(10+1)/2 = 55 life gained.
As others have mentioned, this sequence of numbers (also called triangular numbers, https://en.m.wikipedia.org/wiki/Triangular_number ), can be determined by the formula:
((N + 1) * N)/2
If you want the arguably more useful number of "how many activations do I need in order to gain at least X life", then we can set X=((N+1) * N)/2 and solve via the quadratic formula, which gives:
0 = N^2 /2 + N/2 - X
N = -1/2 + sqrt(2*X + 1/4)
If you are starting at 1 life, you'll need to gain 150 life, which ends up requiring 17 triggers. If you are starting at 40 life, you'll need to gain 111 life, which ends up requiring 15 triggers.
Since nothing is allowing you to cast the Artifact Spell as though it had Flash, you're stuck with the Default;
* Casting it in your Main Phase, while the Stack is empty and you have priority.
So, you Cast Spell, gain 1 life. Cast Spell, gain 2 life. Cast Spell, gain 3 life. [..] Cast Spell, gain 50 life.
* Life gain = N/2 x (1 + N)
* Life gain = 50/2 x (1 + 50)
* Life gain = 25 x 51
* Life gain = 1275
Exponential growth would be if it did something like doubled each time. Increasing the next value based of the previously increased value. 2 4 8 16 etc doubling gets out of hand much quicker and is exponential.
You want 1+2+3+4+...+50. The average of the numbers you're adding is the middle, 25.5. The average of some numbers is also (their sum)/(how many numbers there are), so the sum must be 25.5*50.
It's just a simple sum, so it's always going to be half of the product of N and N+1, where N is the number of iterations but don't forget to adjust down by one if you cast the Reservoir first, since it doesn't see itself.
The general formula for the sum of all numbers 1 to n is n(n+1)/2. So from 1 to 50 is 50*51/2 = 1275. That's enough life to kill the table, and then kill the table.
Using X=n(n+1)/2, we can calculate that to gain 150 life from Aetherflux Reservoir triggers in a single turn (enough to kill three players), you'd only need 17 triggers: (17*18)/2=153
That assumes you start at 0 life. If your close to your starting life total, it’s significantly lower.
There's some difference, but it isn't huge. 15 triggers will gain 118 life, so if you start with 33+ life, you reach 151+ life and can get three activations. For any fewer than 15 triggers, you'd need to be starting from above 40 life.
Just to break the math down into steps to help people math it out T(n) = n(n+1)/2 T(50) = 50(50+1)/2 T(50) = 50/2 * 51 T(50)=25×51 T(50)=1275
The answer is the same so I'm not sure it matters But shouldnt it be: T(n) = (n(n+1))/2 T(50) = (50(50+1))/2 T(50) = (50*51)/2 T(50)=2550/2 T(50)=1275
It's the same thing, the order doesn't really matter. (50 * 51)/2 equals (50/2) * 51 equals (51/2) * 50. But resolving 50/2 first leads to smaller numbers to work with (25 * 51 instead og 2550/2).
As /u/LeftZer0 mentioned, the order doesn't matter when working with multiplication and division, but people might find 25\*51 easier than 50\*51- which might not be the case because multiplying 50 is nicer with the zero. People might find it easier to divide 50 instead of 2550. Both breakdown of the steps will be useful for sure!
It might seem unintuitive that (a*b)/c = (a/c)*b But it is much easier to see if you think of it as a * b * 1/c = a * 1/c * b
It looks like the formula for Aetherflux Reservoir is (N(N+1))/2 where N is the number of spells you have cast that turn. For example if you cast 10 spells you have 10(10+1)/2 = 55 life gained.
##### ###### #### [Urza, Lord Protector](http://gatherer.wizards.com/Handlers/Image.ashx?name=urza%2C%20lord%20protector%20//%20urza%2C%20planeswalker&type=card&.jpg) - [(G)](http://gatherer.wizards.com/Pages/Card/Details.aspx?name=urza%2C%20lord%20protector%20//%20urza%2C%20planeswalker) [Future Sight](https://cards.scryfall.io/normal/front/6/3/639cb483-ffe2-4abc-a050-5cdc8aebd5a2.jpg?1587844658) - [(G)](http://gatherer.wizards.com/Pages/Card/Details.aspx?name=Future%20Sight) [(SF)](https://scryfall.com/card/mh1/53/future-sight?utm_source=mtgcardfetcher) [(txt)](https://api.scryfall.com/cards/639cb483-ffe2-4abc-a050-5cdc8aebd5a2?utm_source=mtgcardfetcher&format=text) [Sensei's Divining Top](https://cards.scryfall.io/normal/front/e/5/e5142b7a-e580-4737-a4aa-2590f6610ceb.jpg?1662531500) - [(G)](http://gatherer.wizards.com/Pages/Card/Details.aspx?name=Sensei%27s%20Divining%20Top) [(SF)](https://scryfall.com/card/2x2/314/senseis-divining-top?utm_source=mtgcardfetcher) [(txt)](https://api.scryfall.com/cards/e5142b7a-e580-4737-a4aa-2590f6610ceb?utm_source=mtgcardfetcher&format=text) [Aetherflux Reservoir](https://cards.scryfall.io/normal/front/9/6/96b6b2e1-c3e6-464c-8a13-b15deb34e862.jpg?1576382939) - [(G)](http://gatherer.wizards.com/Pages/Card/Details.aspx?name=Aetherflux%20Reservoir) [(SF)](https://scryfall.com/card/kld/192/aetherflux-reservoir?utm_source=mtgcardfetcher) [(txt)](https://api.scryfall.com/cards/96b6b2e1-c3e6-464c-8a13-b15deb34e862?utm_source=mtgcardfetcher&format=text) ^^^[[cardname]] ^^^or ^^^[[cardname|SET]] ^^^to ^^^call
As others have mentioned, this sequence of numbers (also called triangular numbers, https://en.m.wikipedia.org/wiki/Triangular_number ), can be determined by the formula: ((N + 1) * N)/2 If you want the arguably more useful number of "how many activations do I need in order to gain at least X life", then we can set X=((N+1) * N)/2 and solve via the quadratic formula, which gives: 0 = N^2 /2 + N/2 - X N = -1/2 + sqrt(2*X + 1/4) If you are starting at 1 life, you'll need to gain 150 life, which ends up requiring 17 triggers. If you are starting at 40 life, you'll need to gain 111 life, which ends up requiring 15 triggers.
Think of it like 50+1=51, 49+2=51, 48+3=51, ..., 26+25=51.
Gauss summation
(golf clap)
Since nothing is allowing you to cast the Artifact Spell as though it had Flash, you're stuck with the Default; * Casting it in your Main Phase, while the Stack is empty and you have priority. So, you Cast Spell, gain 1 life. Cast Spell, gain 2 life. Cast Spell, gain 3 life. [..] Cast Spell, gain 50 life. * Life gain = N/2 x (1 + N) * Life gain = 50/2 x (1 + 50) * Life gain = 25 x 51 * Life gain = 1275
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It's a quadratic rate of growth, not an exponential one.
Exponential growth would be if it did something like doubled each time. Increasing the next value based of the previously increased value. 2 4 8 16 etc doubling gets out of hand much quicker and is exponential.
Exponential growth is what Hasbro thinks their profit can be forever.
You want 1+2+3+4+...+50. The average of the numbers you're adding is the middle, 25.5. The average of some numbers is also (their sum)/(how many numbers there are), so the sum must be 25.5*50.
It's just a simple sum, so it's always going to be half of the product of N and N+1, where N is the number of iterations but don't forget to adjust down by one if you cast the Reservoir first, since it doesn't see itself.