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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 ) ≡ 1 2 + ... + n 2 = n ( n + 1)(2 n + 1) 6 Base case: Lets show that P(1) is true. So let’s put in n = 1 in P(n). L.H.S = 1 2 = 1 R.H.S = 1(1 + 1)(2 × 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. Let’s assume that P(k) is true. We have: P ( k ) ≡ 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 ) ⇒ 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 ) ≡ 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)....
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This note was uploaded on 01/10/2011 for the course EE 100 taught by Professor Razasuleman during the Spring '10 term at University of Engineering & Technology.
 Spring '10
 RazaSuleman

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