Induction I P(O is true · For all n∈NP(n) implies F(n+1) now our job is reduced to proving these two statements. The first is true because P(O)asserts that a sum of zero terms is equal to 0(0+1)/2=0 The second statement is more complicated. But remember the basic plan for prov ing the validity of any implication: assume the statement on the left and then prove the statement on the right. In this case, we assume P(n) 1+2+3+…+nn(n+1) in order to prove P(n+1) +2+3+…+n+(n+1)≈(n+1) These two equations are quite similar; in fact, adding (n+ 1) to both sides of the first equation and simplifying the right side gives the second equation 1+2+3+.…+n+(m+1) n(n+1) +(m+1) (n+2)(n+1 Thus, if P(n)is true, then so is P(n+1). This argument is valid for every natural number m, so this establishes the second fact required by the induction principle. In effect, weve just proved that P(O)implies P(1), P(1)implies P(2), P(2)implies P(3), etc. all in one fell swoop true for all natural n. And so the theorem is proved Ciple says that the predicate P(n)is With these two facts in hand, the induction prind 2.1 A Template for Induction Proofs The proof of Theorem 1 was relatively simple, but even the most complicated induction proof follows exactly the same template. There are five components 1. State that the proof uses induction. This immediately conveys the overall structure of the proof, which helps the reader understand your argument 2. Define an appropriate predicate P(n). The eventual conclusion of the induction argument will be that P(n) is true for all natural n. Thus, you should define the predicate P(n)so that your theorem is equivalent to(or follows from) this conclu sion. Often the predicate can be lifted straight from the claim, above The predicate P(n) is called the"induction hypothesisInduction I 3 • P(0) is true. • For all n ∈ N, P(n) implies P(n + 1). So now our job is reduced to proving these two statements. The first is true because P(0) asserts that a sum of zero terms is equal to 0(0 + 1)/2 = 0. The second statement is more complicated. But remember the basic plan for prov￾ing the validity of any implication: assume the statement on the left and then prove the statement on the right. In this case, we assume P(n): n(n + 1) 1 + 2 + 3 + . . . + n = 2 in order to prove P(n + 1): (n + 1)(n + 2) 1 + 2 + 3 + . . . + n + (n + 1) = 2 These two equations are quite similar; in fact, adding (n + 1) to both sides of the first equation and simplifying the right side gives the second equation: n(n + 1) 1 + 2 + 3 + . . . + n + (n + 1) = + (n + 1) 2 (n + 2)(n + 1) = 2 Thus, if P(n) is true, then so is P(n + 1). This argument is valid for every natural number n, so this establishes the second fact required by the induction principle. In effect, we’ve just proved that P(0) implies P(1), P(1) implies P(2), P(2) implies P(3), etc. all in one fell swoop. With these two facts in hand, the induction principle says that the predicate P(n) is true for all natural n. And so the theorem is proved! 2.1 A Template for Induction Proofs The proof of Theorem 1 was relatively simple, but even the most complicated induction proof follows exactly the same template. There are five components: 1. State that the proof uses induction. This immediately conveys the overall structure of the proof, which helps the reader understand your argument. 2. Define an appropriate predicate P(n). The eventual conclusion of the induction argument will be that P(n) is true for all natural n. Thus, you should define the predicate P(n) so that your theorem is equivalent to (or follows from) this conclu￾sion. Often the predicate can be lifted straight from the claim, as in the example above. The predicate P(n) is called the “induction hypothesis