1條簡單A.maths 數,我唔知錯咩...10分!!

[匿名]

2007-09-08 2:23 am

唔駛俾其它ans 我,我識用其它方法答岩,
只不過唔明,咁樣計,點解錯:
(條題目冇錯,大家唔駛質疑)
Prove each of the following by mathematics induction,where n is a natural number:
唔夠位,跳步先=.=....
Assume S(k) is true
1/(2k-1)(2k+1)=k/2k+1
When n=k+1,
L.H.S.= 1/(2k-1)(2k+1)+1/[2(k+1)-1][2(k+1)+1]
=k/(2k+1)+1/(2k+1)(2k+3)
=[k(2k+1)(2k+3)+1(2k+1)]/(2k+1)^2*(2k+3)--->交差雙乘法
=(k+1)(2k+1)(2k+3)/(2k+1)^2*(2k+3)
=(k+1)/(2k+1)
R.H.S.=(k+1)/[2(k+1)+1]
=(k+1)/2k+3
L.H.S. =/=R.H.S?!

回答 (2)

NCY

2007-09-14 9:02 am

✔ 最佳答案

你應該是通錯分母

Assume S(k) is true
1/(2k-1)(2k+1)=k/2k+ 1
When n=k+1,
L.H.S.
= 1/(2k-1)(2k+1)+1/[2(k+1)-1][2(k+1)+1]
=k/(2k+1)+1/(2k+1)(2k+3) ←這部係岩
=k(2k+3)/(2k+1)(2k+3)+1/(2k+1)(2 k+3) ←但之後要通分母
=k(2k+3) +1/(2k+1)(2k+3) ←但之後要通分母
=(2k2+3k+1)/(2k+1)(2k+3)
= (2k+1)(k+1)/(2k+1)(2k+3)
= (k+1)/(2k+3) ←約去(2k+1)
=R.H.S.
∴By the principle of Mathematical Induction,S(n) is true for all natural numbers n.

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用戶53314

2007-09-08 2:41 am

L.H.S is

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