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October 13th, 2018, 08:05 PM   #1
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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)
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October 13th, 2018, 10:04 PM   #2
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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
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October 14th, 2018, 12:53 AM   #3
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Why is it 4S(n)=2^2+....+(4n-2)^2?
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October 14th, 2018, 01:53 AM   #4
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4S(n) = 2²(1)² + . . . + 2²(2n - 1)² = 2² + . . . + (4n-2)²
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