M135F08A3S

M135F08A3S - MATH 135 Assignment #3 Solutions Hand-In...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
MATH 135 Fall 2008 Assignment #3 Solutions Hand-In Problems 1. (a) ± 18 0 ² = 18! 0!18! = 1 (b) ± 19 17 ² = 19! 17!2! = 19 · 18 2 · 1 = 171 (c) ± 9 3 ² = 9! 3!6! = 9 · 8 · 7 3 · 2 · 1 = 84 (d) ± 12 5 ² = 12! 5!7! = 12 · 11 · 10 · 9 · 8 5 · 4 · 3 · 2 · 1 = 11 · 9 · 8 = 792 (e) Using the Binomial Theorem, ( x 2 - 2 y ) 4 = ( x 2 ) 4 + ± 4 1 ² ( x 2 ) 3 ( - 2 y ) + ± 4 2 ² ( x 2 ) 2 ( - 2 y ) 2 + ± 4 3 ² ( x 2 )( - 2 y ) 3 + ( - 2 y ) 4 = x 8 - 8 x 6 y + 24 x 4 y 2 - 32 x 2 y 3 + 16 y 4 (f) The term containing x 9 is ± 5 2 ² (2 x 3 ) 3 ( - 5) 2 = 2000 x 9 , so the coefficient is 2000. 2. We prove the result by strong induction on n . Base Cases When n = 1, we have 3 · 2 n - ( - 3) n = 6 - ( - 3) = 9 = x 1 . When n = 2, we have 3 · 2 n - ( - 3) n = 12 - 9 = 3 = x 2 . Therefore, the result holds when n = 1 and n = 2. Induction Hypothesis Suppose that the result holds for n = 1 , 2 , . . . , k , for some k P , k 2. That is, suppose that x n = 3 · 2 n - ( - 3) n for n = 1 , 2 , . . . , k . Induction Conclusion Consider n = k + 1. Then x k +1 = - x k + 6 x k - 1 = - (3 · 2 k - ( - 3) k ) + 6(3 · 2 k - 1 - ( - 3) k - 1 ) (by Induction Hypothesis) = - 3 · 2 k + 18 · 2 k - 1 + ( - 3) k - 6( - 3) k - 1 = 2 k - 1 [ - 6 + 18] + ( - 3) k - 1 [ - 3 - 6] = 2 k - 1 [12] + ( - 3) k - 1 [ - 9] = 3 · 2 k +1 - ( - 3) k +1 so the result holds for n = k + 1. Therefore, the result holds for all n P by POSI.
Background image of page 1

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

View Full DocumentRight Arrow Icon
3. We prove the result by strong induction on n . Base Cases When n = 1, we have 1 2 (2 n + ( - 2) n ) = 1 2 (2 - 2) = 0 = y 1 . When n = 2, we have 1 2 (2 n + ( - 2) n ) = 1 2 (4 + 4) = 4 = y 2 . Therefore, the result holds when n = 1 and n = 2. Induction Hypothesis Suppose that the result holds for n = 1 , 2 , . . . , k , for some k P , k 2. That is, suppose that y n = 1 2 (2 n + ( - 2) n ) for n = 1 , 2 , . . . , k . Induction Conclusion Consider n = k + 1. Then y k +1 = 4 y k - 1 = 4 ( 1 2 (2 k - 1 + ( - 2) k - 1 ) ) (by Induction Hypothesis) = 1 2 (2 2 )(2 k - 1 + ( - 2) k - 1 ) = 1 2 (2 2 2 k - 1 + ( - 2) 2 ( - 2) k - 1 ) = 1 2 (2 k +1 + ( - 2) k +1 ) so the result holds for n = k + 1. Therefore, the result holds for all n P by POSI. 4. We prove the result by strong induction on n . Base Cases When n = 1, we have 2( - 1) n - ( - 2) n + 3 n = 2( - 1) - ( - 2) + 3 = 3 = t 1 . When n = 2, we have 2( - 1) n - ( - 2) n + 3 n = 2(1) - 4 + 9 = 7 = t 2 . When n = 3, we have 2( - 1) n - ( - 2) n + 3 n = 2( - 1) - ( - 8) + 27 = 33 = t 3 . Induction Hypothesis Suppose the result is true for n = 1 , 2 , . . . , k , for some k P , k 3. That is, suppose t n = 2( - 1) n - ( - 2) n + 3 n for n = 1 , 2 , . . . , k . Induction Conclusion Consider n = k + 1. Then t k +1 = 7 t k - 1 + 6 t k - 2 = 7(2( - 1) k - 1 - ( - 2) k - 1 + 3 k - 1 ) + 6(2( - 1) k - 2 - ( - 2) k - 2 + 3 k - 2 ) = 14( - 1) k - 1 - 7( - 2) k - 1 + 7(3 k - 1 ) + 12( - 1) k - 2 - 6( - 2) k - 2 + 6(3 k - 2 ) = ( - 1) k - 2 [14( - 1) + 12] - ( - 2) k - 2 [7( - 2) + 6] + 3 k - 2 [7(3) + 6] = ( - 1) k - 2 [ - 2] - ( -
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 11/01/2008 for the course MATH 135 taught by Professor Andrewchilds during the Fall '08 term at Waterloo.

Page1 / 8

M135F08A3S - MATH 135 Assignment #3 Solutions Hand-In...

This preview shows document pages 1 - 3. Sign up to view the full document.

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