1/x(x+3)+1/(x+3)(x+6)+...+1/(x+15)(x+18)
=1/3[1/x-1/(x+3)]+1/3[1/(x+3)-1/(x+5)]+……+1/3[1/(x+15)-1/(x+18)]
=1/3[1/x-1/(x+3)+1/(x+3)-1/(x+5)+……+1/(x+15)-1/(x+18)]
=1/3[1/x-1/(x+18)]
=1/3[(x+18)/x(x+18)-x/x(x+18)]
=1/3[18/x(x+18)]
=6/x(x+18)
1/x(x+3)+1/(x+3)(x+6)+...+1/(x+15)(x+18)
2个回答
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