Abraham P Hillman's Algebra through problem solving PDF

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T - 1. Then the values of a1 and d would determine the values of all the terms. A geometric progression b1, ... , bt is one for which there is a fixed number r such that bn+1 = bnr for n = 1, 2, ... , t - 1; its terms are determined by b1 and r. It is not surprising that mathematical induction is very useful in proving results concerning quantities that are defined inductively, however, it is sometimes necessary or convenient to use an alternate principle, called strong mathematical induction.

35. Prove the following property of the Fibonacci numbers: j n j'0 n (&1)j Fs%2n&2j ' Fs%n. j * 36 Prove an analogue of the formula of Problem 35 for the Lucas numbers. 37 Find a compact expression, without using the sigma notation, for 1@n % 2(n & 1) % 3(n & 2) % þ % (n & 1)@2 % n@1, that is, for j (k % 1)(n & k). n&1 k'0 50 Chapter 7 COMBINATIONS AND PERMUTATIONS We have seen in the previous chapter that (a + b)n can be written as n n n n&1 n n&k k n n a % a b%þ% a b %þ% b 0 1 k n where we have the specific formula for the binomial coefficients: n k ' n!

5@218 (b) Express the nth term similarly. (c) Sum the first five hundred terms. * 44. In the sequence 1, 2, 3, 6, 7, 14, 15, 30, 31, ... a term in an even numbered position is double the previous term, and a term in an odd numbered position (after the first term) is one more than the previous term. (a) What is the millionth term of this sequence? (b) Express the sum of the first million terms compactly. 28 Chapter 5 MATHEMATICAL INDUCTION In mathematics, as in science, there are two general methods by which we can arrive at new results.

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Algebra through problem solving by Abraham P Hillman


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