How can I find n to make the sum of a sequence equal a specific number?

[lanedance]

notice it is always an odd term, if N is the number of terms and S is the 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


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

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

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

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

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

 

[Denyven]

Kk, I've done some work, and I think I found a pattern to the sum.

\frac{\sqrt{2n+1}-1}{2}

 

[um0123]

thanks lanedance, and phase shifter, and to everyone who helped.

together denny and i got the answer, and we couldn't ever do it without you guys.

 

[PhaseShifter]

Kk, I've done some work, and I think I found a pattern to the sum.

\frac{\sqrt{2n+1}-1}{2}

Right.

 

[Denyven]

SWEET, thanks guys, I got the answer. All your help, from everyone, was very much appreciated.