Algebra through problem solving - download pdf or read online

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By Abraham P Hillman

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5! Similarly, 8 2 ' 8! 6! 8 4 ' 8! 4! n r This leads us to conjecture that n! holds in all cases. (n & r)! ' mathematical induction in the following theorem. THEOREM: If n and r are integers with 0 # r # n, then n r ' n! (n & r)! Proof: If n = 0, the only allowable value of r is 0 and 41 0 0 = 1. Since n! 0! (n & r)! 0! the formula holds for n = 0. Now let us assume that it holds for n = k. Then k r&1 ' k! (k & r % 1)! k r ' k! (k & r)! Using (1), above, we now have k%1 k k k! k! (k & r % 1)! (k & r)!

2407. (c) The first eighty terms of -3, -7, -11, -15, ... (d) The first n terms of the arithmetic progression a, a + d, a + 2d, ... 9. Find the fourth, seventh, and ninth terms of the geometric progression with first term 2 and ratio 3. 10. Find the fourth and sixth terms of the geometric progression with first term 2 and ratio -3. 11. Find the next three terms of the geometric progression 2, 14, 98, ... 12. Find the next three terms of the geometric progression 6, -2, 2/3, -2/9, ... 13. Find both possible values of x if 7, x, 252 are three consecutive terms of a geometric progression.

Thus a (k + 1)-sided polygon has a total of k(k&3) % (k & 2) % 1 diagonals. But: 2 k(k&3) % (k & 2) % 1 2 k 2 &3k % 2k & 2 ' 2 30 k2 & k & 2 2 (k % 1)(k & 2) ' 2 (k % 1)[(k % 1) &3] ' 2 ' This is the desired formula for n = k + 1. So, by assuming that the formula Dn = n(n - 3)/2 is true for n = k, we have been able to show it true for n = k + 1. This, in addition to the fact that it is true for n = 3, proves that it is true for all integers greater than or equal to 3. ) The method of mathematical induction is based on something that may be considered one of the axioms for the positive integers: If a set S contains 1, and if, whenever S contains an integer k, S contains the next integer k + 1, then S contains all the positive integers.

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Algebra through problem solving by Abraham P Hillman

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