Discrete Mathematics with Graph Theory (3rd Edition) 136

Discrete Mathematics with Graph Theory (3rd Edition) 136 -...

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134 Solutions to Exercises of k + 1 terms equals the sum of k terms plus the (k + l)st term. The induction hypothesis, together with the fact that ak+1 = a + kd says that this is equal to k (k(k-1) ) 2[2a + (k -l)d) + a + kd = ak + a + 2 + k d =ak+a+ (k(k;l))d= (k;1)(2a+kd) as required. By the Principle of Mathematical Induction, the result is true for all n ~ 1. 23. Let the sequence be a, a + d, a + 2d, . . .. We are given 1 2 0 (2a + 9d) = 2 ~ (2a + 5d), that is, 20 lOa + 45d = 12a + 30d, or 2a = 15d. Also, "2(2a + 19d) = 1360, so 20a + 190d = 1360. Substituting 20a = 10(2a) = 150d, we obtain 340d = 1360, so d = 4 and a = 125d = 30. The sum of n terms of the sequence is ~[60 + 4(n - 1)) = ~(4n + 56) = 2n 2 + 28n. 24. Let x, y, z, w be a, a+d, a+ 2d, a+3d, respectively. We are given that4a+6d = 8, so d = ~(2 -a). We are also given that a( a + 3d) + (a + d) (a + 2d) = - 2, which is a 2 + 3ad + d 2 = -1. Substituting d = ~(2 -a), we obtain a 2 +2(2 -a)a+ ~(4 -4a+a 2 ) = -1, so a 2 -4a- 5
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