MAT102H5 Quiz: Solutions-PSG

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2 Jan 2019
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MAT102H5 Full Course Notes
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Mat102 - introduction to mathematical proofs - utm - fall 2018. Solutions to selected problems from problem set g. Xk=1 k( 1)k = 1 + 2 3 + 4 5 + 6 . + 99) + (2 + 4 + 6 + . 4. 6. 11 (a) base case: if n = 1, then. Inductive step: assume that the statement holds for n = k, that is, k. 4. 6. 15(a) assume that 23k 1 + 5 3k is divisible by 11. Let us show that 23(k+2) 1 + 5 3k+2 is also divisible by. 23(k+2) 1 + 5 3k+2 = 23k 1 26 + 5 3k. 9 = 23k 1 64 + 5 3k. 4. 6. 17 base case: for n = 1 we have 2 13 2 1 + 3 = 3 = a1. Inductive step: assume that ak = 2k3 2k + 3 for some k. consider ak+1 = ak + 6k(k + 1) = 2k3 2k + 3 + 6k(k + 1)

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