Data Structures & Alogs HW_Part_23

Data Structures & Alogs HW_Part_23 - 89 14.3 From...

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89 14.3 From Equation 2.2 we know that n X i =1 i 2 = 2 n 3 +3 n 2 + n 6 . Thus, when summing the range a i b ,weget b X i = a i 2 = 2 b 3 +3 b 2 + b 6 2 a 3 +3 a 2 + a 6 = 2( b 3 a 3 )+3( b 2 a 2 )+( b a ) 6 . 14.4 We need to do some rearranging of the summation to get something to work with. Start with n X i =1 i 2 = n X i =1 ( i +1 1) 2 . Substituting i for i 1 ,weget n X i =1 i 2 = n 1 X i =0 ( i +1) 2 = n 1 X i =0 ( i 2 +2 i +1) . The i 2 terms mostly cancel, leaving n 2 = n 1 X i =0 (2 i +1) =2 n 1 X i =0 i + n. n 2 n 2 = n 1 X i =0 i Substituting back i 1 for i ,weget n X i =1 = n 2 + n 2 .
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90 Chap. 14 Analysis Techniques 14.5 F ( n ) = 2+4+ ··· +2 n 2 F ( n ) = 4+8+ ··· +2 n +1 When we subtract, we get 2 F ( n ) F ( n )= F ( n )=2 n +1 2 . Thus, n X i =1 2 i =2 n +1 2 . 14.6 Call our summation G ( n ) , then G ( n )= n X i =1 i 2 n i =2 n 1 +2 2 n 2 +3 2 n 3 + ··· + n 2 0 . 2 G ( n )=2 n X i =1 i 2 n i =2 n +2 2 n 1 +3 2 n 2 + ··· + n 2 1 . Subtracting, we get 2 G ( n ) G ( n )= G ( n )=2 n +2 n 1 +2 n 2 + ··· +2 1 n 2 0 . This is simply
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This document was uploaded on 10/31/2011 for the course BCN 3431 at University of Florida.

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Data Structures & Alogs HW_Part_23 - 89 14.3 From...

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