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 October 13th, 2018, 08:05 PM #1 Newbie   Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 Mathematical Induction Q, part a) Prove by mathematical induction that 1^2 + 3^2 +...+(2n-1)^2 = 1/3(n)(2n-1)(2n+1) for all positive integers n. part b) Hence, find the value of 110^2 + 114^2 + 118^2 +...+218^2 + 222^2. I have done part a), but I have no idea how to start part b)
 October 13th, 2018, 10:04 PM #2 Global Moderator   Joined: Dec 2006 Posts: 20,303 Thanks: 1974 Let S(n) = 1/3(n)(2n - 1)(2n + 1), so that S(1) = 1. S(n) - S(n - 1) = 1/3(n)(2n - 1)(2n + 1) - 1/3(n - 1)(2n - 3)(2n - 1) $\hspace{99px}$ = 1/3(2n - 1)(n(2n + 1) - (n - 1)(2n - 3)) $\hspace{99px}$ = (2n - 1)² 4S(n) = 2² + . . . + (4n - 2)² 110² + 114² + 118² + . . . + 218² + 222² = 4S(56) - 4S(27) $\hspace{284px}$ = (4/3(56)(111)(113) - 4/3(27)(53)(55)) $\hspace{284px}$ = 4/3(702408 - 78705) $\hspace{284px}$ = 831604 Thanks from hy2000
 October 14th, 2018, 12:53 AM #3 Newbie   Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 Why is it 4S(n)=2^2+....+(4n-2)^2?
 October 14th, 2018, 01:53 AM #4 Global Moderator   Joined: Dec 2006 Posts: 20,303 Thanks: 1974 4S(n) = 2²(1)² + . . . + 2²(2n - 1)² = 2² + . . . + (4n-2)²

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