MATHEMATICAL INDUCTION~~~~急!!

2008-09-03 7:26 am
prove, by mathematical induction,that the following propositions are true
1^2/1x3+2^2/3x5+3^2/5x7+....+n^2/(2n-1)(2n+1)=n(n+1)/2(2n+1)

回答 (1)

2008-09-03 7:55 am
✔ 最佳答案
Let P(n) be the proposition'1^2/1x3+2^2/3x5+3^2/5x7+....+n^2/(2n-1)(2n+1)=n(n+1)/2(2n+1)'

When n=1
   LHS=1^2/(2X1-1)(2X1+1)
      =1/3
   RHS=1(1+1)/2(2X1+1)
      =2/6
      =1/3
  ∴LHS=RHS
  ∴P(1)is true.
Assume that P(k) is true for some positive integer k.
i.e.,1^2/1x3+2^2/3x5+3^2/5x7+....+k^2/(2k-1)(2k+1)=k(k+1)/2(2k+1)
When n=1
   LHS=1^2/1x3+2^2/3x5+3^2/5x7+....+k^2/(2k-1)(2k+1)+(k+1)^2/[2(k+1)-1][2(k+1)+1]
      =k(k+1)/2(2k+1)+(k+1)(k+1)/(2k+1)(2k+3)
      =[k(k+1)(2k+3)+2(k+1)(k+1)]/2(2k+1)(2k+3)
      =(k+1)(2k^2+5k+2)/2(2k+1)(2k+3)
      =(k+1)(2k+1)(k+2)/2(2k+1)(2k+3)
      =(k+1)(k+2)/2(2k+3)
   RHS=(k+1)[(k+1)+1]/2[2(k+1)+1)
      =(k+1)(k+2)/2(2k+3)
  ∴LHS=RHS
  ∴P(K+1)is true.
By the principle of mathematical induction,P(n) is true for all positive intgers n.
參考: 自己


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