Download e-book for iPad: Algebra through problem solving by Abraham P Hillman

By Abraham P Hillman

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13. Guess a formula for each of the following and prove it by mathematical induction: (a) 1 1 1 1 % % % ... % . (x 2 % y 2 ). 14. Guess a formula for each of the following and prove it by mathematical induction: (a) 1@2 % 2@3 % 3@4 % ... % n(n % 1). (b) 1 1 1 1 % % % ... % . 1@3 3@5 5@7 (2n & 1)(2n % 1) 15. Guess a simple expression for the following and prove it by mathematical induction: 1 & 1 22 1 & 1 32 1 & 1 42 þ 1 & 1 n2 . 16. Find a simple expression for the product in Problem 15, using the factorization x 2 & y 2 ' (x & y)(x % y).

N 5) ' 4(1 % 2 % 3 % ... % n)3. 6. (15 % 25 % 35 % ... % n 5) % (17 % 27 % 37 % ... % n 7) ' 2(1 % 2 % 3 % ... % n)4 35 7. 3n % 7n & 2 is an integral multiple of 8. * 8. 2@7n % 3@5n & 5 is an integral multiple of 24. * 9. x 2n & y 2n has x + y as a factor. 10. x 2n%1 % y 2n%1 has x % y as a factor. 11. For all integers n, prove the following: (a) 2n3 + 3n2 + n is an integral multiple of 6. (b) n5 - 5n3 + 4n is an integral multiple of 120. 12. Prove that n(n2 - 1)(3n + 2) is an integral multiple of 24 for all integers n.

Now the identity 2 m m % 2 1 ' 2 m(m & 1) % m ' m2 & m % m ' m2 2 holds for all integers m, and we can use the formulas 1 1 % 2 1 % 3 1 % ... % n 1 ' n%1 2 1 2 % 2 2 % 3 2 % ... % n 2 ' n%1 3 to show that 44 &2 7 ' &8, 12 % 22 % ... % n 2 ' 2 1 2 % 1 1 % 2 ' 2 1 2 % 2 2 % ... % ' 2 n%1 3 ' 2 % 2 2 % n 2 2 1 % ... % 2 n 2 % 1 1 % ... % % 2 1 % n 1 n 1 n%1 2 (n % 1)n(n & 1) (n % 1)n % 6 2 ' 2n 3 & 2n 3n 2 % 3n % 6 6 ' 2n 3 % 3n 2 % n 6 ' n(n % 1)(2n % 1) . 6 Frequently in mathematical literature a short notation for sums is used which involves the Greek capital letter sigma, written '.

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