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Unformatted text preview: Thus, we can simplify Y n = a ! n 3 + b ! n 2 + c ! n . Now we compute: ! n ({ X i }) = n 2 ! n ({ Y i }) = Y n " Y n " 1 = a # n 3 + b # n 2 + c # n " ( a # ( n " 1) 3 + b # ( n " 1) 2 + c # ( n " 1)) This gives us the equation: ! n ({ X i }) = n 2 = 3 a " n 2 + (2 b # 3 a ) n + ( a # b + c ) = ! n ({ Y i }) For this equality to hold we must have that 3 a = 1 2 b ! 3 a = a ! b + c = These equations can be solved to get a = 1 3 , b = 1 2 , and c = 1 6 . Thus we can conclude that: i 2 i = n ! = 1 3 n 3 + 1 2 n 2 + 1 6 n for all n . Exercise. Find a formula for X n = i 3 i = n ! ....
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This note was uploaded on 07/14/2011 for the course MAC 3472 taught by Professor Jury during the Spring '07 term at University of Florida.
 Spring '07
 JURY
 Calculus, Equations

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