homework_1 - -(11)4 n + (11)4 n-4 n +1 = ((11) n-4 n )(11)...

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Homework 1 1.4. (a) When we evaluate the sum f ( n ) = 1 + 3 + · · · + (2 n - 1) for low values of n we obtain f (1) = 1, f (2) = 4, f (3) = 9 and f (4) = 16. From this data, it seems reasonable to guess that f ( n ) = n 2 . (b) Let P n be the statement “1 + 3 + · · · + (2 n - 1) = n 2 .” We showed that P 1 , P 2 , P 3 and P 4 are true in (a). For the induction step, suppose P n is true so 1 + 3 + · · · + (2 n - 1) = n 2 and add 2 n + 1 = 2( n + 1) - 1 to both sides. We obtain 1 + 3 + · · · + (2 n + 1) = n 2 + (2 n + 1) which is 1 + 3 + · · · + (2( n + 1) - 1) = ( n + 1) 2 so P n +1 is true. We have shown that P n implies P n +1 so the induction step holds, and the principle of mathematical induction implies that P n is true for all n . 1.6. Let P n be the statement “(11) n - 4 n is divisible by 7.” The basis for induction P 1 is true since 11 - 4 = 7. For the induction step, suppose P n is true. To verify P n +1 we write (11) n +1 - 4 n +1 = (11) n +1
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Unformatted text preview: -(11)4 n + (11)4 n-4 n +1 = ((11) n-4 n )(11) + 4 n (11-4) . Since (11) n-4 n is divisible by 7, and 4 n (11-4) = 7(4) n is divisible by 7, we have that (11) n +1-4 n +1 is divisible by 7. Therefore, P n implies P n +1 so the induction step holds, and the principle of mathematical induction implies that P n is true for all n . Polyas Paradox. The induction step does not hold for n = 1. If there are n + 1 = 2 horses, then the induction step asks us to partition { 1 , 2 } into the sets { 1 } and { 2 } which do not overlap. The rst base case for which the induction step works is n = 2, but then the statement of the theorem is obviously false: Two horses can have two dierent colors. 1...
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