201Smt2_1061 - n 0(1 x)n = n 1 n 2 x x2 n n xn(1 xn 1 = 1 x x2 x3 xn(1 x 1 = 1 x x2 x3 = xi(1 x i=0 1 =(1 x)n n 0 n i = i=0 i=0 1 =(1 x)n n 0 = i=0

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(1 + x ) n = ± n 0 ! + ± n 1 ! x + ± n 2 ! x 2 + ··· + ± n n ! x n (1 - x n +1 ) (1 - x ) = 1 + x + x 2 + x 3 + ··· + x n 1 (1 - x ) = 1 + x + x 2 + x 3 ··· = X i =0 x i 1 (1 + x ) n = ± - n 0 ! + ± - n 1 ! x + ± - n 2 ! x 2 + ··· = X i =0 ± - n i ! x i = 1 + ( - 1) ± n + 1 - 1 i ! x + ( - 1) 2 ± n + 2 - 1 2 ! x 2 + ··· = X i =0 ( - 1) i ± n + i - 1 i ! x i 1 (1 - x ) n = ± - n 0 ! + ± - n 1 ! ( - x ) + ± - n 2 ! ( - x ) 2 + ··· = X i =0 ± - n i ! ( - x ) i = 1 + ( - 1) ± n + 1 - 1 i ! ( - x ) + ( - 1) 2 ± n + 2 - 1 2 ! ( - x ) 2 + ··· = X i =0 ± n + i - 1 i ! x i 1
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March 8, 2006. 50 minutes Total marks: 55. Marks are indicated by ( ). (1) (15) Consider the following recurrence relation; a n +6 a n - 1 +14 a n - 2 +16 a n - 3 +8 a n - 4 = 0 , a 0 = - 1 , a 1 = 1 , a 2 = 2 , a 3 = 3 Write down the general solution and the equations that will determine the unknown coefficients, but do not solve for the coefficients . 2
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This note was uploaded on 10/01/2011 for the course MACM 201 taught by Professor Marnimishna during the Spring '09 term at Simon Fraser.

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201Smt2_1061 - n 0(1 x)n = n 1 n 2 x x2 n n xn(1 xn 1 = 1 x x2 x3 xn(1 x 1 = 1 x x2 x3 = xi(1 x i=0 1 =(1 x)n n 0 n i = i=0 i=0 1 =(1 x)n n 0 = i=0

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