# Induction step now we show that whenever pk is true

• Notes
• 4

This preview shows pages 2–4. Sign up to view the full content.

Induction Step: Now we show that whenever P(k) is true, P(k+1) is true i.e P ( k ) P ( k + 1). We have: P ( n ) 1 3 + ... + k 3 = (1 + ... + k ) 2 Let’s add ( n + 1) 3 to both sides of P(k). L.H.S = 1 3 + ... + n 3 + ( n + 1) 3 = R.H.S = (1 + ... + n ) 2 + ( n + 1) 3 Now, we know that 1 + ... + n = n X i =1 i = n ( n + 1) 2 Therefore, (1 + ... + n ) 2 = n 2 ( n + 1) 2 2 2 Then, (1 + ... + n ) 2 + ( n + 1) 3 = n 2 ( n + 1) 2 4 + ( n + 1) 3 = ( n + 1) 2 [ n 2 4 + ( n + 1)] = ( n + 1) 2 4 [ n 2 + 4 n + 4] = ( n + 1) 2 ( n + 2) 2 4 = ( n + 1)[( n + 1) + 1] 2 2 This last expression we can see, from our formula of n i =1 to be equal to: = " n +1 X i =1 i # 2 = [1 + ... + n + ( n + 1)] 2 = R.H.S of P(k+1) 2

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

So, again P ( k ) P ( k + 1) and P(1) is true. Hence, by induction P(n) is true. 2(a) n X i =1 (2 i - 1) = 2 n X i =1 ( i ) - n X i =1 (1) = 2 n (1 + n ) 2 - n = n 2 2(b) n X i =1 (2 i - 1) 2 = n X i =1 (4 i 2 - 4 i + 1) = 4 n X i =1 ( i 2 ) - 4 n X i =1 ( i ) + n X i =1 (1) Using n i =1 i 2 = n ( n +1)(2 n +1) 6 from Question 1(a), = 4 n ( n + 1)(2 n + 1) 6 - 4 n (1 + n ) 2 + n = 8 n 3 + 4 n 2 + 8 n 2 + 4 n 6 - 12 n - 12 n 2 6 + 6 n 6 = 8 n 3 - 2 n 6 = 4 n 3 - n 3 3 The problem is in the proof of P ( K ) = P ( k + 1) When there is one student the statement is certainly true. However, when we have two students. We have a problem. The argument in the class was that say if you have say three 1 students A, B and C. If you send out C, then by the induction hypothesis A and B have the same birthday. If you send out A, then B and C have the same birthday. Hence all three have the same birthday. However, this argument breaks down, if you consider only two students A and B. If you send out A, B has one birthday. If you
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