A.maths MI－rlz87tm45t的部落格

標題:

A.maths MI

發問:

a)Prove by mathematical induction that 1^3+2^3+...n^3=n^2(n+1)^2/4 for all positive integers n. b)Use the result of (a)to find, in terms of m, (i) 2^3+4^3+...+(2m)^3 where m is a positive integer (ii)1^3+3^3+...+(2m-1)^3 where m is a positive integer (iii)51^3+53^3+55^3+...+99^3

最佳解答:

(a) Let P(n) be the statement: 13 + 23 + ... + n3 = n2(n + 1)2/4 When n = 1, L.H.S. = 1 and R.H.S. = 1 So P(1) is true. Suppose that P(k) is true where k is a positive integer, 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)2/4](k2 + 4k + 4) = [(k + 1)2/4](k + 2)2 = (k + 1)2(k + 2)2/4 So P(k + 1) is also true. By the principle of M.I., P(n) is true for all positive integers n. (b) (i) 23 + 43 + ... + (2m)3 = 23(13 + 23 + ... + m3) = 8m2(m + 1)2/4 = 2m2(m + 1)2 (ii) 13 + 33 + ... + (2m - 1)3 = [13 + 23 + ... + (2m)3] - [23 + 43 + ... + (2m)3] = (2m)2(2m + 1)2/4 - 2m2(m + 1)2 = m2(2m + 1)2 - 2m2(m + 1)2 = m2[(2m + 1)2 - 2(m + 1)2] = m2[4m2 + 4m + 1 - 2(m2 + 2m + 1)] = m2(2m2 - 1) (iii) 513 + 533 + ... + 993 = (13 + 33 + ... + 993) - (13 + 33 + ... + 493) From (ii) result: 13 + 33 + ... + 993 = 502[2(50)2 - 1] 13 + 33 + ... + 493 = 252[2(25)2 - 1] So: 513 + 533 + ... + 993 = 502[2(50)2 - 1] - 252[2(25)2 - 1] = 2500 x 4999 - 625 x 1249 = 11716875