x(x+3)分之1+(x+3)(x+6)分之1+(x+6)(x+9)分之1+(x+6)(x+9)分之1+...+(x+9

1个回答

  • x(x+3)分之1+(x+3)(x+6)分之1+(x+6)(x+9)分之1+(x+6)(x+9)分之1+...+(x+99)(x+102)分之1

    =1/3 [x(x+3)分之3+(x+3)(x+6)分之3+(x+6)(x+9)分之3+(x+6)(x+9)分之3+...+(x+99)(x+102)分之3]

    =1/3[1/x-1/(x+3)+1/(x+3)-1/(x+6)+1/(x+6)-1/(x+9)+……+1/(x+99)-1/(x+102)]

    =1/3[1/x-1/(x+102)]

    =1/3×102/x(x+102)

    =34/(x²+102x)

    看其中的一个3/(x+3)(x+6)=[(x+6)-(x+3)]/(x+3)(x+6)

    =(x+6)/(x+3)(x+6)-(x+3)/(x+3)(x+6)

    =1/(x+3)-1/(x+6)

    这个方法称为裂顶法