/sci/ - Science & Math

is it (c)?

>>9364312
Is it?

>>9364322
idk i am a brainlet too

>>9364325
Oh, it's ok then friend

>>9364303

its not 1 at least
its not 2 ether , since thta makes them 1/2 + 1/2 + 1/2 forever

my bet is d) makes it into some geometric series that becomes 1 maybe

>>9364327
nah i am pretty sure it is (c). That is a sum of an infinite geometric series. The common ratio is 2/x and first term is 1/x. put in duh a/1-r and equate it to 1. masths 2 herd 4 me i go sleep now

>>9364347
Well, fuck. My guess was e)

[eqn]\sum_{k=1}^{+\infty} \frac{2^{k-1}}{x^k} = \sum_{k=0}^{+\infty} \frac{2^{k}}{x^{k+1}} = \sum_{k=0}^{+\infty} \frac{1}{2}\frac{2^{k+1}}{x^{k+1}}
= \frac{1}{2}\sum_{k=0}^{+\infty} {(\frac{2}{x})}^{k+1}
= \frac{1}{2} * (\sum_{k=0}^{+\infty}(\frac{2}{x})^k-1 )[/eqn]
[eqn]\Leftrightarrow\frac{1}{2} * (\frac{1}{1-\frac{2}{x}}-1)= 1[/eqn]
[eqn]\Leftrightarrow x= 3[/eqn]

>>9364303
You put in 2 and you get infinite sum of constant values.

1/2 + 1/4 + 1/8 ... is obviously 1. 1/4 + 1/8 + 1/16 is obviously 1/2 and that would be the same exact series as d). Therefore you are only left with c).

2 is too big (you get 1/2 + 1/2 + 1/2) and 4 is too small (you get 1/4 + 1/8 + 1/16 + ...) . So the right answer is 3.

Basically what >>9364374 said

>>9364303
Once you recognize this:
[eqn]\frac{d}{dx} \frac{1}{1-x} = \sum_{n=1}^\infty nx^{n-1} [/eqn]

Then you can work out the derivative, plug in and multiply by x to get what you need.

plot twist it's -1/12

>>9364462
No. 42. 42 is always the correct answer