# Algebra through problem solving by Abraham P Hillman By Abraham P Hillman

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Extra info for Algebra through problem solving

Example text

Find numbers a and b such that n 3 ' 6 n n n %a %b 3 2 1 49 for all integers n. 29. Find numbers r, s, and t such that n4 = n(n - 1)(n - 2)(n - 3) + rn(n - 1)(n - 2) + sn(n - 1) + tn for n = 1, 2, and 3. Using these values of r, s, and t, show that n 4 ' 24 n n n n % 6r % 2s %t 4 3 2 1 for all integers n. 30. Find numbers a, b, c, and d such that n n n n n %a %b %c %d . 5 4 3 2 1 n 5 ' 5! 31. Express j k 4 as a polynomial in n. n k'1 32. Express j k 5 as a polynomial in n. n k'1 33. We define a sequence S0, S1, S2, ...

14. Given that n = a + b + c + d and that a, b, c, and d are non-negative integers, show that n a n&a b n&a&b c n&a&b&c d 15. Express j [a % (k & 1)d] as a polynomial in n. n k'1 n 16. Express A (2k) compactly without using the A notation. k'1 n n&1 k'1 j'0 17. Show that A a k ' A aj%1. 18. Show that j bk ' j bi&2. n&2 n k'1 i'3 48 ' n! d! 19. Evaluate j ai j bi 2 2 i'1 i'1 n A ai 20. Show that i'1 n A bi i'1 and j (aib i ) and show that they are not always equal. 2 i'1 n ' A (aibi ). i'1 21. Prove by mathematical induction that j n s%i s i'0 22.

We then wish to show that (2) (k % 1)[(k % 1)2 % 5] ' 6s with s an integer. Simplifying the difference between the left-hand sides of (2) and (1), we obtain (3) (k % 1)[(k % 1)2 % 5] & k(k 2 % 5) ' 3k(k % 1) % 6. Since k and k + 1 are consecutive integers, one of them is even. Then their product k(k + 1) is even, and may be written as 2t, with t an integer. Now (3) becomes (4) (k % 1)[(k % 1)2 % 5] & k(k 2 % 5) ' 6t % 6 ' 6(t % 1). Transposing, we have (k % 1)[(k % 1)2 % 5] ' k(k 2 % 5) % 6(t % 1).