midsolns

# midsolns - Math 239 Midterm Questions Tuesday June 28 1 Use...

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

Tuesday, June 28 Math 239 Midterm Questions 1. Use the Binomial Theorem to prove that: (a) n X k = 0 ˆ n k ! = 2 n (b) n X k = 0 k ˆ n k ! = n 2 n - 1 Solution: (a) According to the Binomial Theorem, n X k = 0 ˆ n k ! x k = (1 + x ) n . (1) Let x = 1, we get n k = 0 ( n k ) = 2 n . (b) Differentiate both sides of Equation 1 to get n X k = 0 ˆ n k ! kx k - 1 = n (1 + x ) n - 1 . (2) Let x = 1, we get n k = 0 ( n k ) k = n 2 n - 1 . 2. Find the generating function for the number of compositions of n with an even number of parts, each of which is congruent to 1 modulo 3. Solution: Let A = {3 a + 1 : a N 0 }. Let S = S k 0 A 2 k . Let the weight function on A be α ( r ) = r and let the weight function on S be the sum of the weight of the components. The coefficient of x n in Φ S ( x ) is then the number of compositions of n with an even number of parts, each of which is congruent to 1 modulo 3. We have Φ A ( x ) = x + x 4 +··· = x 1 - x 3 . By the sum and product lemma, Φ S ( x ) = X k 0 Φ A ( x ) 2 k = 1 1 - Φ A ( x ) 2 = (1 - x 3 ) 2 (1 - x 3 ) 2 - x 2 = 1 - 2 x 3 + x 6 1 - x 2 - 2 x 3 + x 6 . The k = 0 term is included since by definition, there is a single composition with 0 parts (which is even).

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

View Full Document
Math 239 Midterm Questions 3. The generating function for the set Ω of binary strings that do contain a copy of 0110 is 1 + x 3 1 - 2 x + x 3 - x 4 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern