Maths(infinite sequence)
回答 (2)
2006-11-28 8:39 pm
✔ 最佳答案
let (-1)^(n-1)*(1/n) be the nth term of the sequencethe common different would be
(-1)^(n-1)*(1/n) - (-1)^(n)*(1/(n+1))
=(-1)^(n-1)*(1/n + 1/(n+1))
2006-11-28 9:05 pm
Let nth term be (-1)^(n+1)．(1/(2n-1))
therefore the common different is
T(n+1)-T(n) = (-1)^(n+2)．(1/(2n+1)) + (-1)^(n+1)．(1/(2n-1)
=(-1)^(n+1)[ 1/(2n-1) - 1/(2n+1))
=(-1)^(n+1)[2/(4n^2-1)]
therefore the common different is
T(n+1)-T(n) = (-1)^(n+2)．(1/(2n+1)) + (-1)^(n+1)．(1/(2n-1)
=(-1)^(n+1)[ 1/(2n-1) - 1/(2n+1))
=(-1)^(n+1)[2/(4n^2-1)]
收錄日期: 2021-04-18 19:54:00
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https://hk.answers.yahoo.com/question/index?qid=20061128000051KK00954