利用分項對消法證明
1/(1×2)+1/(3×4)+1/(5×6)+1/(7×8)+⋯1/((2n-1)×2n)=1/(n+1)+1/(n+2)+⋯1/2n
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分項對消法求證明之問題
2014-01-14 9:35 am
回答 (3)
2014-01-14 4:10 pm
✔ 最佳答案
利用分項對消法證明1/(1*2)+1/(3*4)+1/(5*6)+1/(7*8)+⋯+1/[(2n-1)*2n]=1/(n+1)+1/(n+2)+⋯+1/(2n)
Sol
1/[k*(k+1)]
=[(k+1)-k]/[k*(k+1)]
=(k+1)/[k*(k+1)]-k/[k*(k+1)]
=1/k-1/(k+1)
So
1/(1*2)+1/(3*4)+1/(5*6)+1/(7*8)+⋯+1/[(2n-1)*2n]
=Σ(k=1 to n)_1/[(2k-1)*2k]
=Σ(k=1 to n)_[1/(2k-1)-1/(2k)]
=Σ(k=1 to n)_1/(2k-1)-Σ(k=1 to n)_1/(2k)
=Σ(k=1 to n)_1/(2k-1)+Σ(k=1 to n)_1/(2k)-2Σ(k=1 to n)_1/(2k)
=Σ(k=1 to n)_1/(2k-1)+Σ(k=1 to n)_1/(2k)-Σ(k=1 to n)_1/k
=Σ(k=1 to 2n)_1/k-Σ(k=1 to n)_1/k
=1/(n+1)+1/(n+2)+⋯+1/(2n)
2014-01-15 3:51 am
>這里很不錯ａａａｓｈｏｐｓ。ｃｏｍ老婆很喜歡
吣久倍从
吣久倍从
2014-01-14 3:36 pm
1/[K(K+1)]=1/K-1/(K+1)
左式=(1/1-1/2)+(1/3-1/4)+(1/5-1/6)+…+〔1/(2n-1)-1/2n〕
=1/1-1/2+1/3-1/4+1/5-1/6+…+1/(2n-1)-1/2n
=〔1/1+1/2+1/3+1/4+1/5+1/6+…+1/(2n-1)+1/2n〕-2*〔1/2+1/4+1/6+…+1/2n〕
=〔1/1+1/2+1/3+1/4+1/5+1/6+…+1/(2n-1)+1/2n〕-2*(1/2)〔1/1+1/2+1/3+…+1/n〕
=〔1/1+1/2+1/3+1/4+1/5+1/6+…+1/(2n-1)+1/2n〕-〔1/1+1
2014-01-14 07:37:29 補充：
=〔1/1+1/2+1/3+1/4+1/5+1/6+…+1/(2n-1)+1/2n〕-〔1/1+1/2+1/3+…+1/n〕
=1/(n+1)+1/(n+2)+⋯1/2n
左式=(1/1-1/2)+(1/3-1/4)+(1/5-1/6)+…+〔1/(2n-1)-1/2n〕
=1/1-1/2+1/3-1/4+1/5-1/6+…+1/(2n-1)-1/2n
=〔1/1+1/2+1/3+1/4+1/5+1/6+…+1/(2n-1)+1/2n〕-2*〔1/2+1/4+1/6+…+1/2n〕
=〔1/1+1/2+1/3+1/4+1/5+1/6+…+1/(2n-1)+1/2n〕-2*(1/2)〔1/1+1/2+1/3+…+1/n〕
=〔1/1+1/2+1/3+1/4+1/5+1/6+…+1/(2n-1)+1/2n〕-〔1/1+1
2014-01-14 07:37:29 補充：
=〔1/1+1/2+1/3+1/4+1/5+1/6+…+1/(2n-1)+1/2n〕-〔1/1+1/2+1/3+…+1/n〕
=1/(n+1)+1/(n+2)+⋯1/2n
收錄日期: 2021-05-02 10:45:24
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20140114000015KK00427