Mathematical induction111

(a) Let P(n) be the statement 12 + 22 + 32 + ... + n2 = n(n + 1)(2n + 1)/6
When n = 1, LHS = 1, RHS = 1(2)(3)/6 = 1
Thus P(1) is true.
Suppose that P(k) is true where k is a positive integer, i.e.
12 + 22 + 32 + ... + k2 = k(k + 1)(2k + 1)/6
Adding (k + 1)2 to both sides:
12 + 22 + 32 + ... + k2 + (k + 1)2 = k(k + 1)(2k + 1)/6 + (k + 1)2
= [(k + 1)/6][k(2k + 1) + 6(k + 1)]
= [(k + 1)/6](2k2 + 7k + 6)
= (k + 1)(2k + 3)(k + 2)/6
Thus P(k + 1) is also true. By the principle of MI, P(n) is true for all positive integers n.
(b 1) 22 + 42 + 62 + ... + (2n)2 = 22(12 + 22 + 32 + ... + n2)
= 4n(n + 1)(2n + 1)/6
= 2n(n + 1)(2n + 1)/3
(2) 12 - 22 + 32 - 42 +... - 402 = (12 + 22 + 32 + 42 +... + 402) - 2(22 + 42 + 62 +... + 402)
= 40(41)(81)/6 - 40(21)(41)/3
= 22140 - 11480
= 10660
(3) 1x1 + 2x3 + 3x5 + ... + n(2n-1)
= 2(12 + 22 + 32 + ... + n2) - (1 + 2 + ... + n)
= n(n + 1)(2n + 1)/3 - n(n + 1)/2
= [n(n + 1)/6][2(2n + 1) - 3]
= n(n + 1)(4n - 1)/6

2009-09-27 09:49:36 補充：
Some corrections for b2:
(2) 1^2 - 2^2 + 3^2 - 4^2 +... - 40^2 = (1^2 + 2^2 + 3^2 + 4^2 +... + 40^2) - 2(2^2 + 4^2 + 6^2 +... + 40^2)
= 40(41)(81)/6 - 80(21)(41)/3
= 22140 - 22960
= - 820

(b)(2) is negative.