hw2-1-sol - CS 202: Data Structures and Discrete...

Info iconThis preview shows pages 1–3. 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: CS 202: Data Structures and Discrete Mathematics II Fall 2010 Homework 2-1 (Due September 22 in Class) No Late Submissions 1. (14=3+3+3+5 points) Binomial Theorem: (a) Find the 6th term in the expansion of ( x- 1 x 3 ) 10 . Then find the coefficient of 1 x 14 in this expansion. Solution: By Binomial theorem, x- 1 x 3 10 = 10 X k =0 C (10 ,k ) x k (- 1 /x 3 ) 10- k = 10 X k =0 C (10 ,k )(- 1) 10- k x k x 3(10- k ) = 10 X k =0 C (10 ,k )(- 1) 10- k 1 x 30- 3 k- k The 6th term in the expansion is the term for k = 5 which is C (10 , 5)(- 1) 10- 5 1 x 30- 15- 5 =- 252 x 10 From the above expansion, 1 /x 14 will show up in the term where 30- 3 k- k = 14, which means k = 4. Thus, the coefficient of 1 /x 14 is C (10 , 4)(- 1) 10- 4 = C (10 , 4) = 210 . (b) Use the binomial theorem twice to expand ( x + y + z ) 3 . Solution: ( x + y + z ) 3 = ( x + ( y + z )) 3 = 3 ( y + z ) x 3 + 3 1 ( y + z ) x 2 + 3 2 ( y + z ) 2 x + 3 3 ( y + z ) 3 x = x 3 + 3( y + z ) x 2 + 3( y + z ) 2 x + ( y + z ) 3 = x 3 + 3( y + z ) x 2 + 3 x 2 z y 2 + 2 1 z 1 y 1 + 2 2 z 2 y + 3 z y 3 + 3 1 z 1 y 2 + 3 2 z 2 y 1 + 3 3 z 3 y = x 3 + 3( y + z ) x 2 + 3 x ( y 2 + 2 zy + z 2 ) + ( y 3 + 3 zy 2 + 3 z 2 y + z 3 ) = x 3 + 3 x 2 y + 3 x 2 z + 3 xy 2 + 6 xyz + 3 xz 2 + y 3 + 3 zy 2 + 3 z 2 y + z 3 = x 3 + y 3 + z 3 + 3 x 2 z + 3 xy 2 + 3 xz 2 + 3 zy 2 + 3 z 2 y + 6 xyz 2-1-1 CS 202 Homework 2-1 Due September 22 in Class (c) Use the binomial theorem to prove that C ( n, 0) + C ( n, 1)2 + C ( n, 2)2 2 + + C ( n,n )2 n = 3 n Solution: Notice that 2 + 1 = 3 and 1 k = 1 for any k . Therefore, we can rewrite the equation above as C ( n, 0)1 n + C ( n, 1)2 1 n- 1 + C ( n, 2)2 2 1 n- 2 + + C ( n,n )2 n 1 = (1 + 2) n = 3 n . (d) Prove the identity in (c) using a combinatorial counting argument. Solution: We will show that both sides count the number of strings of length n where there are three choices for each letter (say, a , b , and c ). Clearly, the right hand side counts the number of such strings since each position has three independent choices and there are n positions, so by the rule of product we get 3 n ....
View Full Document

This note was uploaded on 11/22/2010 for the course CS 202 taught by Professor Staff during the Spring '08 term at Ill. Chicago.

Page1 / 5

hw2-1-sol - CS 202: Data Structures and Discrete...

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