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Problem with a sequence

  • Thread starter um0123
  • Start date
  • #26
152
0
hey denny, yea this is the kid that i was working on this for 2 hours with, and it i showed him the physics forums and now hes trapped.

plus hes into partcile physics like me!:smile:

but back to busniess.

the 3rd term is

√5 - √7
_______
-2


add that to the previous summation of:

1 - √5
______
-2

and you get:

1 - √7
______
-2
now lets see that pattern. gimme a minute.
 
Last edited:
  • #27
277
1
okay, i simplify and get

√2n-1 - √2n+1
____________
4n
Try again.

What is [tex](\sqrt{2n-1})^{2}-(\sqrt{2n+1})^{2}[/tex]?
 
  • #28
277
1
I am in the same math class as Um0123, and I currently have got
[tex]\frac{\sqrt{2n-1} - \sqrt{2n+1}}{-2}[/tex]

would squaring the fraction be a step in the right direction?
On the right track, but unfortunately you can't square this to get rid of the radicals for several reasons.
Try calculating that fraction for several consecutive values of n, and see if you notice a pattern.
 
  • #29
152
0
is the pattern:

1 - √(n+2)
________
2

EDIT: DAMNIT, THAT ONLY WORKS FOR ODD TERMS!!!!
 
Last edited:
  • #30
lanedance
Homework Helper
3,304
2
no, try putting in the n values as a check, you will need to keep the squareroot
 
  • #31
lanedance
Homework Helper
3,304
2
notice it is always an odd term, if N is the number of terms and S is teh sum wth N terms:

N = 1,
S(N) = (√3- 1)/2

N = 2
S(N) = (√5- 1)/2

N = 3
S(N) = (√7- 1)/2
...
notice it is always includes squareroot of an odd term, any ideas how to write that in terms of N?

Once you have your general equation, solve it for N, when S(N) = 100


-----------------------------
I have to go now (good luck), but here's another more rigourous way to finish the problem:

to help see the pattern & summarise
[tex]S(N) = \sum_{n = 1}^{N} \frac{1}{\sqrt{2n - 1} + \sqrt{2n + 1}}[/tex]

after rationalising the denominator
[tex]S(N) = \sum_{n = 1}^{N} \frac{1}{2} (\sqrt{2n + 1} - \sqrt{2n - 1} )[/tex]

splitting the series & changing dummy to help with substitution:
[tex]S(N) = \frac{1}{2}( (\sum_{n = 1}^{N} \sqrt{2n + 1}) - (\sum_{m = 1}^{N}\sqrt{2m - 1}))[/tex]

now to go forward form here:
- substitute into the 2nd sum only, m = n +1
- the sum limits of the 2nd sum will now become n = 0 to n = N-1 (previously m =1 to N)

cancelling terms should lead to the same pattern you observe
 
Last edited:
  • #32
19
0
Kk, I've done some work, and I think I found a pattern to the sum.

[tex]\frac{\sqrt{2n+1}-1}{2}[/tex]
 
  • #33
152
0
thanks lanedance, and phase shifter, and to everyone who helped.

together denny and i got the answer, and we couldnt ever do it without you guys.
 
Last edited:
  • #34
277
1
Kk, I've done some work, and I think I found a pattern to the sum.

[tex]\frac{\sqrt{2n+1}-1}{2}[/tex]
Right.
 
  • #35
19
0
SWEET, thanks guys, I got the answer. All your help, from everyone, was very much appreciated.
 

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