what is the sum of 1/(i^2-1) with i from 2 to 20 and remember that the sum of i^2 =(n(n+1)(2n+1))/6
20
∑
i=2 __1 ___
i^2 - 1
2007-01-20 18:22:00 · 9 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
[2...20]∑1/(i^2-1)
= [2...20](1/2) ∑[1/(i-1) - 1/(i+1)]
= (1/2)[1-(1/3)+(1/2)-(1/4)+...-1/21]
= (1/2)[1+(1/2)-(1/20)-(1/21)]
= 589/840
= 0.70119
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I made a small error in the first response.
2007-01-20 18:30:13 · answer #1 · answered by sahsjing 7 · 1⤊ 0⤋
Using previous answerer method (all credit goes to him) that 1/(i^2-1) = (1/(i-1) - 1/(i+1))/2 your sum = (1/2)*(1+1/2+1/3+1/4 ... +1/19 - 1/3-1/4-...-1/19-1/20-1/21) = (1/2)*(1+1/2-1/20-1/21) =589/840 = 0.70119
2007-01-20 19:04:40 · answer #2 · answered by fernando_007 6 · 2⤊ 0⤋
20
∑1/(i^2 - 1) = 0.701190476
i=2
2007-01-20 21:21:11 · answer #3 · answered by Helmut 7 · 0⤊ 0⤋
It's a permutation and combination question.
No one likes doing that crap.
Do your own homework :p
2007-01-20 18:29:41 · answer #4 · answered by p_rutherford2003 5 · 0⤊ 1⤋
I don't think sum is what you're looking for.
2007-01-20 18:54:49 · answer #5 · answered by seksy4life 2 · 0⤊ 1⤋
You're missing some vital info. my friend.
2007-01-20 18:25:28 · answer #6 · answered by mathlete1 3 · 0⤊ 0⤋
figure it out on your own.
2007-01-20 18:30:09 · answer #7 · answered by Anonymous · 0⤊ 0⤋
erm... Do your own homework ok??
2007-01-20 19:09:11 · answer #8 · answered by Eliel S 3 · 0⤊ 1⤋
STOP CHEATING
2007-01-20 18:29:38 · answer #9 · answered by gill73115 3 · 0⤊ 0⤋