1/k(k+1)(k+2)=1/2k+1/(k+1)+1/2-查字典问答网
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  1/k(k+1)(k+2)=1/2k+1/(k+1)+1/2(k+2)的拆法1/k(k+1)(k+2)=1/2k+1/(k+1)+1/2(k+2)是怎么拆出来的?我看书上就一步有什么公式或者定理嘛?

  1/k(k+1)(k+2)=1/2k+1/(k+1)+1/2(k+2)的拆法

  1/k(k+1)(k+2)=1/2k+1/(k+1)+1/2(k+2)是怎么拆出来的?

  我看书上就一步有什么公式或者定理嘛?

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2020-01-3120:08
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黄允华

  你题目有问题吧

  应该是1/k(k+1)(k+2)=1/2k-1/(k+1)+1/2(k+2)

  用待定系数法就可以了

  设1/k(k+1)(k+2)

  =a/k+b/(k+1)+c/(k+2)

  =[a(k+1)(k+2)+b(k+2)k+c(k+1)k]/k(k+1)(k+2)

  =[(a+b+c)k^2+(3a+2b+c)k+2a]/k(k+1)(k+2)

  等式两边相等,则(a+b+c)k^2+(3a+2b+c)k+2a=1

  所以有

  a+b+c=0

  3a+2b+c=0

  2a=1

  a=1/2b=-1c=1/2

2020-01-31 20:09:46
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