This preview shows pages 1–4. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full DocumentThis preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
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...
View
Full
Document
This note was uploaded on 07/12/2008 for the course MAT 243 taught by Professor Callahan during the Spring '06 term at ASU.
 Spring '06
 CALLAHAN
 Math

Click to edit the document details