# MAT102 quiz

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University of Toronto Mississauga

Mathematics

MAT102H5

Shay Fuchs

Fall

Description

MAT102S - Intro. to Mathematical Proofs - Winter 2011 - UTM
Quiz 3 (Version B) - SOLUTIONS
3
1. [4 marks] Prove by induction that for all n 2 N, n ▯ 4n is divisible by 3.
Solution:
3 3
Basic step) For n = 1, n ▯ 4n = 1 ▯ 4 = ▯3 and ▯3 is divisible by 3.
Induction step) Suppose that n ▯4n is divisible by 3. We need to show that then (n+1) ▯4(n+1)
is also divisible by 3. In order to do so, we rearrange the expression (n + 1) ▯ 4(n + 1) as
(n + 1) ▯ 4(n + 1) = n + 3n + 3n + 1 ▯ 4n ▯ 4 = n ▯ 4n + (3n + 3n ▯ 3):
By induction assumption we know that n ▯ 4n is divisible by 3. The term (3n + 3n ▯ 3) is also
divisible by 3. Therefore (n + 1) ▯ 4(n + 1) is divisible by 3.
2. [4 marks] Prove by induction that for all n 2 N,
2 2 2 2 1
1 + 2 + 3 + ▯▯▯ + n = 6n(1 + n)(1 + 2n):
Solution:
2 1
Basic step) For n = 1, 1 = 1 and 6(1)(1 + 1)(1 + 2) = 1 so the formula is TRUE.
Induction step) Suppose that the formula is true for n. Then we have
2 2 2 2 2 1 2
1 + 2 + 3 + ▯▯▯ + n + (n + 1) = 6n(1 + n)(1 + 2n) + (n + 1)
2
n(1 + n)(1 + 2n) + 6(n + 1)
=
6

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