This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Problem Set 1 Calculus 1 Abdullah Khalid, Hassan Bukhari November 10, 2010 1(a) We have to show that the following statement is true. P ( n ) â‰¡ 1 2 + ... + n 2 = n ( n + 1)(2 n + 1) 6 Base case: Lets show that P(1) is true. So letâ€™s put in n = 1 in P(n). L.H.S = 1 2 = 1 R.H.S = 1(1 + 1)(2 Ã— 1 + 1) 6 = 1 So we see tht L.H.S = R.H.S and so P(1) is true. Induction Step: Now we would like to show that whenever P(k) is true, P(k+1) is true. Letâ€™s assume that P(k) is true. We have: P ( k ) â‰¡ 1 2 + ... + k 2 = k ( k + 1)(2 k + 1) 6 Then add ( k + 1) 2 to both sides of P(k). L.H.S = 1 2 + ...k 2 + ( k + 1) 2 = R.H.S = k ( k + 1)(2 k + 1) 6 + ( k + 1) 2 = k ( k + 1)(2 k + 1) + 6( k + 1) 2 6 = ( k + 1)[ k (2 k + 1) + 6( k + 1)] 6 = ( k + 1)(2 k 2 + k + 6 k + 6) 6 = ( k + 1)(2 k 2 + 7 k + 6) 6 = ( k + 1)(2 k 2 + 4 k + 3 k + 6) 6 = ( k + 1)[2 k ( k + 2) + 3( k + 2)] 6 = ( k + 1)( k + 2)(2 k + 3) 6 = k + 1)( k + 2)[2( k + 1) + 1] 6 = R.H.S of P(k+1) 1 So we have shown that P ( k ) â‡’ P ( k + 1) and that P(1) is true. Hence, by induction P(n) is true. 1(b) Now we have to show that P ( n ) â‰¡ 1 3 + ... + n 3 = (1 + ...n ) 2 Base case: We proceed in the same way we did in the part (a). We first prove the base case i.e. P(1) is true.We proceed in the same way we did in the part (a)....
View
Full Document
 Spring '10
 RazaSuleman
 Product Rule, Mathematical Induction, Inductive Reasoning, Mathematical logic, base case

Click to edit the document details