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We could also use a different letter like j or i . The top and bottom of Σ tell us which k ’s we add. The bottom k=1 means we start with k = 1 . When do we stop adding? That is explained by the top n . So, we add all the integers from 1 up to n . That is what n k k=1 59 Blast into Math! Sets of numbers: mathematical playground means. Let’s practice. What does 10 5k k=8 mean? First, what are we adding? This time, we are adding 5 times k . Where do we start? We start with k = 8 . When do we stop?

2 3. Proof: we must use the induction assumption to show that the formula will be true for the next integer, n + 1, in other words, we must show 1 + 2 + . . + n + (n + 1) = (n + 1)(n + 1 + 1) . 2 To do this, we use step 2. We can split the sum on the left side above into the first n numbers and then the last one, (1 + 2 + . . + n) + n + 1. + n = 56 n(n + 1) , 2 Blast into Math! Sets of numbers: mathematical playground so we can substitute: 1 + 2 + . . + n + (n + 1) = n(n + 1) + (n + 1). 2 We can combine these numbers into a single fraction by putting the multiplicative identity in disguise: 2 1= , 2 so we can use this to put n + 1 into disguise n + 1 = 1 ∗ (n + 1) = 2 2(n + 1) ∗ (n + 1) = 2 2 so 1 + 2 + .

Hint for # 3: You can use # 2 to prove # 3. There are other correct but different proofs too! • Hint for # 4: All a, b ∈ N are at least as big as 1 , a ≥ 1, b ≥ 1. So, what you’re trying to do is find some n ∈ N so that when you multiply it with a , you get something BIGGER than b . Remember: N is closed under addition and multiplication. So, starting with a and b , you can add or multiply them together (or with other natural numbers), and you’ll end up with a natural number. Play around with a and b and see if you can use them to build an n which solves the problem.

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