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Unformatted text preview: Math 104: Introduction to Analysis SOLUTIONS Alexander Givental HOMEWORK 1 1.1. Prove that 1 2 + 2 2 + + n 2 = 1 6 n ( n + 1)(2 n + 1) for all n N . Put f ( n ) = n ( n + 1)(2 n + 1) / 6. Then f (1) = 1, i.e the theorem holds true for n = 1. To prove the theorem, it suFces to assume that it holds true for n = m and derive it for n = m + 1, m = 1 , 2 , 3 , ... . We have f ( m + 1) f ( m ) = 1 6 ( m + 1)[(2 m + 3)( m + 2) m (2 m + 1)] = 1 6 ( m + 1)(6 m + 6) = ( m + 1) 2 . By the induction hypothesis, f ( m ) = m k =1 k 2 , and therefore f ( m + 1) = f ( m ) + ( m + 1) 2 = m +1 s k =1 k 2 . 1.9 Decide for which n the inequality 2 n > n 2 holds true, and prove it by mathematical induction. The inequality is false n = 2 , 3 , 4, and holds true for all other n N . Namely, it is true by inspection for n = 1, and the equality 2 4 = 4 2 holds true for n = 4. Thus, to prove the inequality for all n 5, it suFces to prove the following inductive step: or any n 4, if 2 n n 2 , then 2 n +1 > ( n + 1) 2 . This is not hard to see: 2 n +1 = 2 2 n 2 n 2 , which is greater than ( n +1) 2 provided that ( n +1) < 2 n i.e. when n > 1 / ( 2 1) = 2+1, which includes all integers n 4. 1.12bc. Put ( n k ) := n ! /k !( n k )! , prove (a) ( n k ) + ( n k 1 ) = ( n +1 k ) for k = 1 , ..., n , and (b) derive the binomial theorem by induction. (b) Note that 1 k + 1 n k + 1 = ( n k + 1) + k k ( n k + 1) = n + 1 k ( n k + 1) . Therefore n ! k !( n k )! + n ! ( k 1)!( n k + 1)! = n ! ( k 1)!( n k )! b 1 k + 1 n k + 1 B = n ! ( k 1)!( n k )! b n + 1 k ( n k + 1) B = ( n + 1)! k !( n k + 1)! . (c) or n = 1 we have ( a + b ) n = a + b = ( 1 1 ) a + ( 1 1 ) b . 1 2 Suppose for some n 1 ( a + b ) n = n s k =0 p n k P a k b n k . Then ( a + b ) n +1 = ( a + b ) n s l =0 p n l P a l b n l = n +1 s k =0 bp n k 1 P + p n k PB a k b n +1 k = n +1 s k =0 p n + 1 k P a k b n +1 k . 2.5. Show that [3 + 2] 2 / 3 does not represent a rational number. Suppose it does represent a rational number q . Then q 3 = [3+ 2] 2 = 9+6 2+2 = 11+6 2. Then 2 = ( q 3 11) / 6 Q , which contradicts irrationality of 2. HOMEWORK 2 3.6b. Use induction to prove  a 1 + a 2 + + a n   a 1  +  a 2  + +  a n  for n numbers a 1 , a 2 , . . ., a n . 1 For n = 1, we need  a 1   a 1  , which is true tautologically. 2 Suppose the required inequality is true for n = k . Then for n = k + 1, using the triangle inequality, we obtain:  a 1 + + a k + a k +1   a 1 + + a k  +  a k +1   a 1  + +  a k  +  a k +1  ....
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 Spring '09
 Math

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