數學歸納法跪求詳解

(a)
Let P(n) be 1/(1x2x3)+1/(2x3x4)+...+1/n(n+1)(n+2)=n(n+3)/4(n+1)(n+2)
P(1): 1/6=4/24
Assume P(k) is true,then P(k+1):
1/(1x2x3)+...+1/n(n+1)(n+2)+1/(n+1)(n+2)(n+3)=......+1/(n+1)(n+2)(n+3)
RHS:
n(n+3)/4(n+1)(n+2)+1/(n+1)(n+2)(n+3)
=n(n+3)(n+3)/4(n+1)(n+2)(n+3)+4/4(n+1)(n+2)(n+3)
=(n(n+3)(n+3)+4)/4(n+1)(n+2)(n+3)
=(n^3+6n^2+9n+4)/4(n+1)(n+2)(n+3)
=(n+1)(n+1)(n+4)/4(n+1)(n+2)(n+3)
=(n+1)(n+4)/4(n+2)(n+3)
PROOF IS COMPLETED

2010-09-12 22:35:34 補充：
(b)
1/(1×2×3)+1/(2×3×4)+...+1/n(n+1)(n+2)>=6/25
n(n+3)/4(n+1)(n+2)>=6/25
25n(n+3)>=24(n+1)(n+2)
25n^2+75n>=24n^2+72n+48
n^2+3n-48>=0
n>=5.589 or n<=-8.589(rej.)

n=6