mat25-practice-final-sol

# mat25-practice-final-sol - Math 25 Advanced Calculus UC...

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Math 25: Advanced Calculus UC Davis, Spring 2011 Math 25 — Solutions to practice problems Question 1 For n = 0 , 1 , 2 , 3 , . . . and 0 k n define numbers C n k by C n k = n ! k !( n - k )! = n ( n - 1) . . . ( n - k + 1) k ! (for k = 0 and k = n we define C n 0 = C n n = 1). (a) Prove that for all n 1 and 1 k n - 1, C n k = C n - 1 k - 1 + C n - 1 k Proof. C n - 1 k - 1 + C n - 1 k = ( n - 1)( n - 2) . . . ( n - k + 1) ( k - 1)! + ( n - 1)( n - 2) . . . ( n - k ) k ! = k + ( n - k ) k · ( n - 1) . . . ( n - k + 1) ( k - 1)! = n k · ( n - 1) . . . ( n - k + 1) ( k - 1)! = C n k (b) Prove by induction on n that for all n 1, n X k =0 C n k = C n 0 + C n 1 + C n 2 + . . . + C n n = 2 n Proof. For n = 1 we have C n 0 + C n 1 = 1+1 = 2 1 , so the claim is true. Let n 1 be given, and assume the claim is true for that value of n . Then, by using the result of (a) above, we see that n +1 X k =0 C n +1 k = 1 + n X k =1 C n +1 k + 1 = 2 + n X k =1 ( C n k - 1 + C n k ) = 2 + n X j =0 C n j + n X k =1 C n k = 2 + n - 1 X j =0 C n j - 1 ! + n - 1 X k =0 C n k - 1 ! = 2 + (2 n - 1) + (2 n - 1) = 2 · 2 n = 2 n +1 , 1

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where we used the inductive hypothesis in the transition from the second row to the third. This shows that the claim is true for n +1 and completes the induction. Question 2 (a) Let ( a n ) n =1 , ( b n ) n =1 be sequences of real numbers. For each of the follow- ing identities, explain what assumptions are needed to ensure that the identity is valid: i. lim n →∞ ( a n + b n ) = lim n →∞ a n + lim n →∞ b n ii. lim n →∞ ( a n · b n ) = lim n →∞ a n · lim n →∞ b n iii. lim n →∞ a n b n = lim n →∞ a n lim n →∞ b n Solution. By a theorem we learned, i. and ii. are valid under the assump- tion that the sequences ( a n ) n =1 and ( b n ) n =1 are convergent. For iii., one needs the additional assumption that the limit of ( b n ) n =1 is not zero.
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