psol08 - Solutions to Homework 8, Math 55 Section 6.1 4....

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Solutions to Homework 8, Math 55 Section 6.1 4. (a) Since 0 =- 3 0 + 4 0, a n = 0 is obviously a solution. (b) Similarly, since 1 =- 3 1 + 4 1, a n = 1 is a solution. (c) Plugging in, we need to prove (- 4) n =- 3(- 4) n- 1 + (- 4) n- 2 . Factoring, this is equivalent to 16(- 4) n- 2 = 12(- 4) n- 2 + 4(- 4) n- 2 , which is obvious. (d) Since a n = 1 and a n = (- 4) n are solutions by the previous two parts, we calculate 2(- 4) n + 3 = 2[- 3(- 4) n- 1 + 4(- 4) n- 2 ] + 3[- 3 1 + 4 1] =- 3(2(- 4) n- 1 + 3) + 4(2(- 4) n- 2 + 3) . 10. (a) In the n th year, the account will pay 0 . 09 a n- 1 in interest, which added to the starting balance of a n- 1 gives a n = a n- 1 + 0 . 09 a n- 1 = 1 . 09 a n- 1 . We also have that a = $1000. (b) We easily see that a n = (1 . 09) n $1000. (c) From the previous part, a 100 = (1 . 09) 100 $1000 $5 , 529 , 041. 14. (a) We have a n = a n- 1 + 1000 + (0 . 05) a n- 1 = (1 . 05) a n- 1 + 1000. (c) We find that a n =- 20000 is a particular solution to this recurrence relation, and the corresponding homogeneous equation has solution b n = C (1 . 05) n . Thus, a n = C (1 . 05) n- 20000 for some C , and since a = 50000, we see C = 70000, and a n = 70000(1 . 05) n- 20000.. (b) Plugging in n = 8 to the solution from part (c), we get a 8 $83 , 422. 24. (a) The number of such bit strings ending with a 1 is a n- 1 ; the number of such bit strings ending with 10 is a n- 2 ; the number ending with 100 is a n- 3 ; and the number ending with 000 is 2 n- 3 . Therefore, a n = a n- 1 + a n- 2 + a n- 3 +2 n- 3 . (Another possible recurrence relation, gotten by splitting according to whether or not the last three digits are the first occurrence of a triple 0, is a n = 2 a n- 1 +(2 n- 4- a n- 4 ).) (b) We can easily see a 1 = a 2 = 0, and a 3 = 1....
View Full Document

This note was uploaded on 10/10/2009 for the course MATH 55 taught by Professor Strain during the Spring '08 term at University of California, Berkeley.

Page1 / 3

psol08 - Solutions to Homework 8, Math 55 Section 6.1 4....

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online