Math question!

[Quad]

how do find the sum of this series:

sum from n=2 to infinite, n^2/2^n

thankS!

 

[TuxDave]

use maple...hehehe.. sorry... I don't know..

 

[Moonbeam]

2+3+4+5+6+7+8+9+----------->infinity?

What is that n squared divided by 2 to the nth power?

2 squared divided by 2 squared plus 2 cubed divided by 3 squared divided by 2 to the 3rd plus 4 squared divided by 2 to the forth etc? or

4/4+9/8+16/16+25/32+36/64+49/128 etc

Is that your question?

 

[gopunk]

S = 4/2^2 + 9/2^3 + 16/2^4 + 25/2^5 ...

S/2 = 4/2^3 + 9/2^4 + 16/2^5 + 25/2^6...

S - S/2 = S/2 = 1 + 5/2^3 + 7/2^4 + 9/2^5 + ...

S/2/2 = S/4 = 1/2 + 5/2^4 + 7/2^5 + 9/2^6 + ...

S/2 - S/4 = S/4 = 1/2 + 5/2^3 + 2/2^3 + 2/2^4 + 2/2^5 +...

We know that sum of A^i from i to infinity is 1/(1-A). Therefore, S/4 = 1/2 + 5/2^3 + (1/(1-1/2) - 1/2 - 1)

calculate right side and multiply by 4

i get 13/2... my arithmatic somewhere might have been wrong, but you get the idea.

 

[eLiu]

by looking at the first coupla terms, I'm guessing it's past 5 but below 6. Running it on maple shows me that it's 5.5. I really don't know how to do it w/o a computer though. Obviously, it converges, but I only remember two ways of finding the sum of a convergent series: telescoping (where terms cancel in a pattern and you can figure out what the answer will be) and geometric (a/1-r). Beyond that, it starts getting complicated...(like, showing what 1/n^2 converges to...pi^2/6 iirc).

 

[eLiu]

Originally posted by: gopunk
S = 4/2^2 + 9/2^3 + 16/2^4 + 25/2^5 ...

S/2 = 4/2^3 + 9/2^4 + 16/2^5 + 25/2^6...

S - S/2 = S/2 = 1 + 5/2^3 + 7/2^4 + 9/2^5 + ...

S/2/2 = S/4 = 1/2 + 5/2^4 + 7/2^5 + 9/2^6 + ...

S/2 - S/4 = S/4 = 1/2 + 5/2^3 + 2/2^3 + 2/2^4 + 2/2^5 +...

We know that sum of A^i from i to infinity is 1/(1-A). Therefore, S/4 = 1/2 + 5/2^3 + (1/(1-1/2) - 1/2 - 1)

calculate right side and multiply by 4

i get 13/2... my arithmatic somewhere might have been wrong, but you get the idea.

Wow, cool...I'll hafta keep this method in mind for next time I get a problem like this.