一條f.4 數學歸納法問題

[匿名]

2007-07-16 6:36 am

我有一題數唔識呀...我依家係自學緊a maths....
prove by mathematical induction that...
1/(2*5*8)+1/(5*8*11)+1/(8*11*14)+.......+1/((3n-1)(3n+2)(3n+5))=(1/60)+1/(6(3n+2)(3n+5))
for all natural number n
唔該各位幫我諗下...我諗唔到n=k+1之後可以點先同r.h.s.一樣，唔該各位

回答 (1)

lotrinity

2007-07-16 6:53 am

✔ 最佳答案

我諗(1/60)+1/(6(3n+2)(3n+5))應改為))=(1/60)-1/(6(3n+2)(3n+5))
when n=1
1/((3n-1)(3n+2)(3n+5))
=1/80
=1/60-1/240
=RHS

it is assumed to be true when n = k

when n=k+1
1/(2*5*8)+1/(5*8*11)+1/(8*11*14)+.......+1/((3n-1)(3n+2)(3n+5))
=1/(2*5*8)+1/(5*8*11)+1/(8*11*14)+.......+1/((3k-1)(3k+2)(3k+5))+1/((3k+2)(3k+5)(3k+8))
=(1/60) - 1/(6(3kl+2)(3k+5))+1/((3k+2)(3k+5)(3k+8))
=(1/60) - (1/6-1/(3k+8))(1/(3k+2)(3k+5)
=(1/60) - (3k+2)/6(3k+8)) * (1/(3k+2)(3k+5))
=(1/60) - 1/6(3k+5)(3k+8)
=(1/60) - 1/6(3n+2)(3n+5)
=RHS
By the principle of mathmatical induction ,it is true for all natural no. n

2007-07-15 22:54:08 補充：
你係咪都係f.3 ga

2007-07-17 23:03:19 補充：
冇咩方法都係死做爛做MI太簡單

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