# 230hw6 - d dx of (1 + x ) n ?) Problem 4. Note that we can...

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Math 230 E Fall 2011 Homework 6 Drew Armstrong Problem 1. Without using Fermat’s little Theorem , prove the Freshman’s Binomial Theorem : For all a,b,p Z with p prime we have ( a + b ) p a p + b p mod p. (Hint: For 1 < k < p show that p divides the numerator of ( p k ) = p ! k !( p - k )! but not the denominator; hence p divides ( p k ) .) Problem 2. Recreate Euler’s (1736) proof of Fermat’s little Theorem. That is, ﬁx a prime p Z and use induction on n to show that n p n mod p for all integers n 0. (Hint: Freshman’s Binomial Theorem.) [This is why I didn’t let you use Fermat’s little Theorem to prove the Freshman’s Binomial Theorem; because I wanted you to use the Freshman’s Binomial Theorem to prove Fermat’s little Theorem!] Problem 3. Use the binomial theorem to prove the following: (a) ( n 0 ) + ( n 1 ) + ( n 2 ) + ··· + ( n n ) = 2 n for all n 1. (b) ( n 0 ) - ( n 1 ) + ( n 2 ) - ··· + ( - 1) n ( n n ) = 0 for all n 1. (c) 0 ( n 0 ) + 1 ( n 1 ) + 2 ( n 2 ) + ··· + n ( n n ) = n 2 n - 1 for all n 1. (Hint: The proofs are one-liners. What is the derivative
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Unformatted text preview: d dx of (1 + x ) n ?) Problem 4. Note that we can write ± n k ² = n ! k !( n-k )! = ( n ) k k ! , where ( n ) k := n ( n-1) ··· ( n-( k-1)). Why would we do this? Because the expression ( z ) k makes sense for any positive integer k and any complex number z ∈ C . Thus we can deﬁne ( z k ) := ( z ) k /k ! for any k ∈ N and z ∈ C . Prove that for all n,k ∈ N we have ±-n k ² = (-1) k ± n + k-1 k ² . Problem 5. Let x,z ∈ C be complex numbers with | x | < 1. Newton’s Generalized Binomial Theorem says that (1 + x ) z = 1 + ± z 1 ² x + ± z 2 ² x 2 + ± z 3 ² x 3 + ··· where the right hand side is a convergent inﬁnite series. Use this to obtain an inﬁnite series expansion of (1 + x )-2 when | x | < 1. (Hint: Apply Problem 4.)...
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## This note was uploaded on 01/08/2012 for the course MATH 461 taught by Professor Armstrong during the Fall '11 term at University of Miami.

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