Pre-Calculus Pre-Calculus Math Forum

 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: 21,113 Thanks: 2327 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: 21,113 Thanks: 2327 4S(n) = 2²(1)² + . . . + 2²(2n - 1)² = 2² + . . . + (4n-2)² Tags induction, mathematical Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post ahmed786 Algebra 2 May 5th, 2017 10:48 AM spuncerj Pre-Calculus 1 November 28th, 2014 05:43 PM medos Algebra 5 October 31st, 2012 04:54 PM MathematicallyObtuse Algebra 2 February 5th, 2011 04:59 PM remeday86 Applied Math 1 June 20th, 2010 10:52 AM

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