1/X(X+3)+1/(X+3)(X+6)+……＋1/(X+99)(X+102)
=1/3[1/x-1/(x+3)]+1/3[1/(x+3)-1/(x+6)]+...+1/3[1/(x+99)-1/(x+102)]
=1/3[1/x-1/(x+102)]
=1/3*101/x(x+102)
=101/3x(x+102)
如果本题有什么不明白可以追问,
求解 1/X(X+3)+1/(X+5)(X+6)+……＋1/(X+99)(X+100)
3个回答
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