M2 Mathematical Induction

15a) Let P(n) be the statement 12 x 4 + 22 x 5 + ... + n2 (n + 3) = n(n + 1)(n2 + 5n + 2)/4

When n = 1, LHS = 4 and RHS = (1/4) x 2 x (12 + 5 + 2) = 4

So P(1) is true.

Suppose that P(k) is true for some positive integer k, i.e.

12 x 4 + 22 x 5 + ... + k2 (k + 3) = k(k + 1)(k2 + 5k + 2)/4

Adding (k + 1)2 (k + 4) to both sides:

12 x 4 + 22 x 5 + ... + (k + 1)2 (k + 4) = k(k + 1)(k2 + 5k + 2)/4 + (k + 1)2 (k + 4)

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

= [(k + 1)/4] (k3 + 9k2 + 22k + 16)

= [(k + 1)/4] (k + 2)(k2 + 7k + 8)

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

Hence P(k + 1) is also true and then P(n) is true for all positive integers n.

b) 12 x 2 + 22 x 3 + ... + n2 (n + 1) = [12 x 4 + 22 x 5 + ... + n2 (n + 3)] - 2(12 + 22 + ... + n2)

= n(n + 1)(n2 + 5n + 2)/4 - n(n + 1)(2n + 1)/3

= (1/12)n(n + 1)[3(n2 + 5n + 2) + 4(2n + 1)]

= (1/12)n(n + 1)(3n2 + 23n + 10)

16a) Let P(n) be the statement 13 + 23 + ... + n3 = n2 (n + 1)2/4

When n = 1, LHS = 1 and RHS = (1/4) x 1 x 22 = 1

So P(1) is true.

Assume that P(k) is true for some positive integer k, i.e.

13 + 23 + ... + k3 = k2 (k + 1)2/4

Adding (k + 1)3 to both sides:

13 + 23 + ... + (k + 1)3 = k2 (k + 1)2/4 + (k + 1)3

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

= (k + 1)2(k + 2)2/4

Hence P(k + 1) is also true and then P(n) is true for all positive integers n.

b) (13 - 1) + (23 - 2) + ... + (n3 - n)

= n2 (n + 1)2/4 - n(n + 1)/2

= [n(n + 1)/4] [n(n + 1) - 2]

= [n(n + 1)/4] (n2 + n - 2)

= n(n -- 2)(n + 1)2/4

15b錯左少少...
最後應該系 n(n+1)[3(n^2+5n+2) - 4(2n+1)]/12
入面個個應該系減號離

2012-04-13 16:04:40 補充：
16b 最尾個factorize都錯左...
n^2 + n -2 = (n-1)(n+2)

所以最後答案: n(n+1)(n-1)(n+2)/4

bcoz im not clever enough, would u pls. show 1or 2 more steps for these steps?

b) 1^2 x 2 + 2^2 x 3 + ... + n^2 (n + 1) = [1^2 x 4 + 2^2 x 5 + ... + n^2 (n + 3)] - 2(1^2 + 2^2 + ... + n^2)

b) (1^3 - 1) + (2^3 - 2) + ... + (n^3 - n)

= n^2 (n + 1)2^/4 - n(n + 1)/2