109 prove that 1 2 2 2 2 n 1 2 n 1 for all n n p n is

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10.9 Prove that 1 + 2 + 2 2 + · · · + 2 n - 1 = 2 n - 1 for all n N . P ( n ) is the statement: 1 + 2 + 2 2 + · · · + 2 n - 1 = 2 n - 1. P (1) is the statement: 2 0 = 1 which is true. Suppose that 1 + 2 + 2 2 + · · · + 2 k - 1 = 2 k - 1. Then 1 + 2 + 2 2 + · · · + 2 k - 1 + 2 k = 2 k - 1 + 2 k = 2 · 2 k - 1 = 2 k +1 - 1 and so P ( k + 1) is true. Therefore, P ( n ) is true for all n by induction.
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2 10.13 Prove that 5 2 n - 1 is a multiple of 8 for all n N . P ( n ) is the statement: 5 2 n - 1 is a multiple of 8. P (1) is the statement: 5 2 - 1 is a multiple of 8, which is true. Suppose that 5 2 k - 1 is a multiple of 8. Say 5 2 k - 1 = 8 m where m is an integer. Then 5 2( k +1) - 1 = 5 2 k · 25 - 1 = 25(5 2 k - 1) + 24 = 25 · 8 m - 24 = 8(25 m - 3). Since 25 m - 3 is an integer, 5 2( k +1) - 1 is a multiple of 8 and so P ( k + 1) is true. Therefore, P ( n ) is true for all n by induction. 10.22 Use induction to prove Bernoulli’s inequality: If 1 + x > 0, then (1 + x ) n 1 + nx for all n in N . P ( n ) is the statement: If 1 + x > 0, then (1 + x ) n 1 + nx . If n = 1, then (1 + x ) 1 = 1 + x 1 + 1 · x and so P (1) is true. Suppose that P ( k ) is true. Then (1 + x ) k 1 + kx . Multiplying both sides by the positive number 1 + x we get (1 + x ) k +1 (1 + kx )(1 + x ). Since x is a real number, x 2 0. Therefore x + x 2 x . Dividing both sides by 1+ x we see that x x/ (1+ x ). Multiplying by k and then adding 1 to both sides we see that 1+ kx 1+ kx 1+ x . Multiplying bot sides by 1 + x we see that (1 + kx )(1 + x ) 1 + x + kx = 1 + ( k + 1) x . Therefore, (1 + x ) k +1 (1 + kx )(1 + x ) 1 + ( k + 1) x and so P ( k + 1) is true. Therefore, P ( n ) is true for all n by induction.
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