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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....
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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.
 Spring '08
 STRAIN
 Math

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