(1/21-1/20)+(1/22-1/21)+(1/23-1/22)+…+(1/2004-1/2003)=

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(1/21-1/20)+(1/22-1/21)+(1/23-1/22)+…+(1/2004-1/2003)=(1/21-1/20)+(1/22-1/21)+(1/23-1/22)+…+(1/2004-

(1/21-1/20)+(1/22-1/21)+(1/23-1/22)+…+(1/2004-1/2003)=
(1/21-1/20)+(1/22-1/21)+(1/23-1/22)+…+(1/2004-1/2003)=

(1/21-1/20)+(1/22-1/21)+(1/23-1/22)+…+(1/2004-1/2003)=
(1/21-1/20)+(1/22-1/21)+(1/23-1/22)+…+(1/2004-1/2003)
=-1/20+1/21-1/21+1/22-1/22+1/23-…-1/2002+1/2003-1/2003+1/2004
=1/2004-1/20
=-124/2505

老大,看列式可知:
式中第1和4 3和6 5和8 7和10 ....n和n+3项相加等于0.
那么我们留下的项目有第2 相倒数第2项.
所以原式=-1/20+1/2004=-124/2505