sols17

# sols17 - k ± k x k − 1 y n − k = n x y n − 1 Now...

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42 17. October 27th: The binomial theorem. Dedekind cuts, I. 17.1 . Explain why adding up every other number on a line of Pascal’s triangle gives a power of 2. For example, from the line for n =6 : 1+0+15+0+15+0+1=32=2 5 . (Use the binomial theorem!) Answer: Plugging x = y =1intheb inom ia ltheoremg ives ° n 0 ± + ° n 1 ± + ° n 2 ± + ° n 3 ± + ··· =2 n as observed in Remark 17.2. Plugging x =1 , y = 1g ives 0 n =(1 1) n = n ² k =0 ° n k ± 1 k ( 1) n k = ° n 0 ± ° n 1 ± + ° n 2 ± ° n 3 ± + ··· Adding up these two expressions shows that 2 · ° n 0 ± +0 · ° n 1 ± +2 · ° n 2 ± +0 · ° n 3 ± + ··· =2 n +0 n =2 n for n> 0. (For n =0 ,0 0 =1 .) D iv id ingby2 ,weobta inthat ° n 0 ± + ° n 2 ± + ° n 4 ± + ··· =2 n 1 for n> 0, explaining the pattern noticed in the statement. ° 17.2 . Prove that n ² k =0 k ° n k ± = n · 2 n 1 . (Hint: Derivative.) Answer: The binomial theorem tells us that n ² k =0 ° n k ± x k y n k =( x

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Unformatted text preview: k ± k x k − 1 y n − k = n ( x + y ) n − 1 . Now plug-in x = y = 1 to get n ² k =0 ° n k ± k 1 k − 1 1 n − k = n (1 + 1) n − 1 = n · 2 n − 1 , which is precisely the statement (since 1 k − 1 1 n − k = 1 for all k and n ). ° 17. OCTOBER 27TH: THE BINOMIAL THEOREM. DEDEKIND CUTS, I. 43 17.3 . Prove that the function Q → D deFned in § 17.5 is injective: that is, prove that if q 1 ° = q 2 , then { x ∈ Q | x < q 1 } ° = { x ∈ Q | x < q 2 } . Answer. Without loss of generality we may assume q 1 < q 2 . Then q 1 °∈ { x ∈ Q | x < q 1 } (since q 1 ° < q 1 ), but q 1 ∈ { x ∈ Q | x < q 2 } (since q 1 < q 2 ). Thus the sets di±er, since they do not have the same elements. °...
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## This note was uploaded on 12/14/2011 for the course MGF 3301 taught by Professor Aluffi during the Fall '11 term at FSU.

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sols17 - k ± k x k − 1 y n − k = n x y n − 1 Now...

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