mathematic induction

First,assume that n is any +ve intergers

Let P(n) be the proposition.

For P(1),L.H.S.=1/10 R.H.S.=1/(6+4)=1/10
so P(1) is true

assume that P(k) is true for all +ve intergers,
i.e.1/(2*5)+1/(5*8)+...............+1/(3k-1)(3k+2)=k/6k+4

For P(k+1),1/(2*5)+1/(5*8)+...............+1/(3k-1)(3k+2)+1/(3k+2)(3k+5)
=k/(6k+4)+1/(3k+2)(3k+5)
=(k(3k+5)+2)/2(3k+2)(3k+5)
=(3k^2+5k+2)/2(3k+2)(3k+5)
=(3k+2)(k+1)/2(3k+2)(3k+5)=(k+1)/(6k+10)=(k+1)/(6(k+1)+4)

so when P(k) is true,P(k+1) is also true

so P(n) is true for all +ve intergers,

Let P(n)be 1/(2*5)+1/(5*8)+...............+1/(3n-1)(3n+2)=n/(6n+4) is true for all natural nos. n
When n=1
LHS=1/(2*5)=1/10
RHS=1/10=LHS
∴P(1)is true
Assume that P(k)is true, where k is a natural no.
ie. 1/(2*5)+1/(5*8)+...............+1/(3k-1)(3k+2)=k/(6k+4) is true, where k is a natural no.
For n=k+1
LHS=1/(2*5)+1/(5*8)+...............+1/(3k-1)(3k+2)+1/(3k+2)(3k+5)
=k/(6k+4)+1/(3k+2)(3k+5)
=k/(2)(3k+2)+1/(3k+2)(3k+5)
=[k/2+1/(3k+5)]/(3k+2)
={[k(3k+5)+2]/2(3k+5)}/(3k+2)
=[(3k^2+5k+2)/2(3k+5)]/(3k+2)
=(3k^2+5k+2)/2(3k+5)(3k+2)
=(3k+2)(k+1)/2(3k+5)(3k+2)
=(k+1)/2(3k+5)
=(k+1)/(6k+10)
=(k+1)/[6(k+1)+4]
=RHS
∴P(k+1)is true
By the principle of M.I., P(n)is true for all natural nos. n

2007-08-24 15:05:13 補充：
最好將D steps寫落張紙到會易睇好多~~