mathematical induction5.5

(a) Sn = [8(1 + 2 + 3 + ... + n)]/[8(1 + 2 + 3 + ... + n) + 1]
= [4n(n + 1)]/[4n(n + 1) + 1]
= 1 - 1/[4n(n + 1) + 1]
(b) Let P(n) be the statement Sn = 1 - 1/[4n(n + 1) + 1]
When n = 1, L.H.S. = 8/9, R.H.S. = 8/9
So P(1) is true.
Suppose that P(k) is true where k is a positive integer, i.e.
(8 x 1)/(12 x 32) + (8 x 2)/(32 x 52) + ... + 8k/[(2k - 1)2(2k + 1)2] = 1 - 1/[4k(k + 1) + 1]
Adding 8(k + 1)/[(2k + 1)2(2k + 3)2] to both sides:
(8 x 1)/(12 x 32) + (8 x 2)/(32 x 52) + ... + 8(k + 1)/[(2k + 1)2(2k + 3)2] = 1 - 1/[4k(k + 1) + 1] + 8(k + 1)/[(2k + 1)2(2k + 3)2]
Consider 1/[4k(k + 1) + 1] - 1/[4(k + 1)(k + 2) + 1]:
1/[4k(k + 1) + 1] - 1/[4(k + 1)(k + 2) + 1] = [4(k + 1)(k + 2) + 1 - 4k(k + 1) - 1]/{[4k(k + 1) + 1][4(k + 1)(k + 2) + 1]}
= (4k2 + 12k + 8 - 4k2 - 4k)/[(4k2 + 4k + 1)(4k2 + 12k + 9)]
= (8k + 8)/[(2k + 1)2(2k + 3)2]
= 8(k + 1)/[(2k + 1)2(2k + 3)2]
Thus,
1 - 1/[4k(k + 1) + 1] + 8(k + 1)/[(2k + 1)2(2k + 3)2] = 1 - 1/[4(k + 1)(k + 2) + 1]
So P(k + 1) is also true. By the principle of MI, P(n) is true for all positive integers n.

答案好似唔岩喎
答案係[(2n+1)^2-1]/[(2n+1)^2]