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Unformatted text preview: 2 Problem 3: Use mathematical induction to show that 5 divides n 5n whenever n is a nonnegative integer. Answer: First we check the smallest case, with n = 0. Then n 5n = 0, which is divisible by 5. Now we assume that n 5n is divisible by 5. Then ( n + 1) 5( n + 1) = n 5 + 5 n 4 + 10 n 3 + 10 n 2 + 5 n + 1n1 = ( n 5n ) + 5 n 4 + 10 n 3 + 10 n 2 + 5 n = ( n 5n ) + 5( n 4 + 2 n 3 + 2 n 2 + n ) . The frst part is a multiple o± 5 by our inductive hypothesis, and the second part is obviously a multiple o± 5, so the whole thing is a multiple o± 5. 3 Problem 4: Use the geometric sum formula to evaluate 2 + 4 3 + 8 9 + ··· + 2 p 2 3 P 8 . Answer: This is 2 + 2 · p 2 3 P + 2 · p 2 3 P 2 + ··· + 2 p 2 3 P 8 = 2(2 / 3) 92 (2 / 3)1 = 38342 6561 ≈ 5 . 84393 . 4...
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 Spring '06
 CALLAHAN
 Math, Prime number, Professor Callahan

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