Astrophysics
  • Astrophysics
How exactly can I tell a series with different index equals another series, could I just plug in numbers?
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Astrophysics
  • Astrophysics
For example I was working on a DE yesterday, but I was not exactly sure why this is true, \[\sum_{n=2}^{\infty} a_nn(n-1)x^{n} = \sum_{n=0}^{\infty} a_nn(n-1)x^{n} \] it's more the a_n that confuses me.
Astrophysics
  • Astrophysics
@amistre64 @jim_thompson5910 @IrishBoy123
IrishBoy123
  • IrishBoy123
astro i personally would just plug them in. i am not sure, however, that that is how you are "supposed" to do it. hope this finds you well 😊

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

Empty
  • Empty
You can see that everything is the same, except for two missing terms of n=0 and n=1, \[\sum_{n=2}^{\infty} a_nn(n-1)x^{n} = \sum_{n=0}^{\infty} a_nn(n-1)x^{n} \] So to prove that, you can pull out the first 2 terms of the sum on the right: \[\sum_{n=2}^{\infty} a_nn(n-1)x^{n} = a_0 0(0-1) + a_1(1(1-1))+ \sum_{n=2}^{\infty} a_nn(n-1)x^{n} \] Now wait, these two sums are the same, so we can remove them by subtracting from both sides: \[0= a_0 0(0-1) + a_1(1(1-1))\] Ok, so now we're left with this, which is I think a pretty clearly true statement, so it's all true.
Astrophysics
  • Astrophysics
Oooooh I think I get it, so we can start at n = 0 because when you get to n=2 it's just the same sum, as the n= 0 and n = 1 go to 0 anyways, ahh I hope I said that correctly! Thanks man! Thanks to @IrishBoy123 as well
Empty
  • Empty
Yeah exactly, those two contributing terms are just 0, so adding 0 never changed nuffin' :P
Astrophysics
  • Astrophysics
That makes so much sense haha
Empty
  • Empty
Don't say that, this is diffy Q it's not supposed to make sense r u outta ur mind?!
Astrophysics
  • Astrophysics
|dw:1446682442924:dw|
anonymous
  • anonymous
@AloneS

Looking for something else?

Not the answer you are looking for? Search for more explanations.