Mathematical Induction 2

Let P(n) be 1/(1*3*5)+1/(3*5*7)+1/(5*7*9)+...+1/[(2n-1)(2n+1)(2n+3)] = [n(n+2)]/[3(2n+1)(2n+3)] for all positive integer n.

When n = 1

LHS

= 1/1x3x5

= 1/15

RHS

= (1) ((1)+2)/3(2(1)+1)(2(1) +3)

= (1)(3)/ 3(3)(5)

= 1/15

As LHS = RHS

So P(1) is true.

Assume P(k) is true, i.e.

1/(1*3*5) + 1/(3*5*7) + 1/(5*7*9) + ... + 1/[(2k-1)(2k+1)(2k+3)] = [k(k+2)]/[3(2k+1)(2k+3)] for some positive integer k.

When n = k + 1,

LHS

= 1/(1*3*5)+1/(3*5*7)+1/(5*7*9)+...+1/[(2n-1)(2n+1)(2n+3)] + 1/(2k+1)(2k+3)(2k+5)

= k(k+2)]/[3(2k+1)(2k+3)] + 3/3(2k+1)(2k+3)(2k+5) 【By assumsion of P(k)】

= [k(k+2) (2k+5) + 3]/ 3(2k+1)(2k+3)(2k+5)

= [(k2+2k)(2k+5) + 3]/ 3(2k+1)(2k+3)(2k+5)

= [2k3 + 9k2 + 10k + 3]/ 3(2k+1)(2k+3)(2k+5)

= (2k+1)[k2+4k+3]/ 3(2k+1)(2k+3)(2k+5)

= (2k+1)(k+1)(k+3)/ 3(2k+1)(2k+3)(2k+5)

= (k+1)(k+3)/ 3(2k+3)(2k+5)

= (k+1)[(k+1)+2]/3[2(k+1)+1][2(k+1)+3]

= RHS

So P(k+1) is true.

By Mathematical Induction, 1/(1*3*5)+1/(3*5*7)+1/(5*7*9)+...+1/[(2n-1)(2n+1)(2n+3)] = [n(n+2)]/[3(2n+1)(2n+3)] is true for all positive integer n.

2007-01-06 16:56:46 補充：
回應你的補充：可以寫得更詳細的，如下：2k³ + 9k² + 10^k + 3= 2(k³ + 3k² + 3k + 1) + 3k² + 4k + 1= 2(k+1)³ + 3(k² + 2k + 1) - 2k - 2= 2(k+1)³ + 3(k+1)² - 2(k+1)= (k+1) [2(k+1)² + 3(k+1) - 2]= (k+1) [(k+1) + 2][2(k+1) - 1]= (k+1)(k+3)(2k + 1)