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 +...+(2n1)^2 = 1/3(n)(2n1)(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: 19,977 Thanks: 1851 
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 
October 14th, 2018, 12:53 AM  #3 
Newbie Joined: Aug 2018 From: HK Posts: 17 Thanks: 0 
Why is it 4S(n)=2^2+....+(4n2)^2?

October 14th, 2018, 01:53 AM  #4 
Global Moderator Joined: Dec 2006 Posts: 19,977 Thanks: 1851 
4S(n) = 2²(1)² + . . . + 2²(2n  1)² = 2² + . . . + (4n2)² 

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induction, mathematical 
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