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Unformatted text preview: 12 2 P n =1 Sn n S1 =1+2+3++n= Sn =1 m n(n + 1) 2 m m *ª¸ B « ( m =1+2+3++n= P Sm 1(1 + 1) =1 2 n(n + 1) 2 1 =1+2+3++m= m(m + 1) 2 P m S m +1 = m *ª´H w E · Ð • n(n + 1) m 2 =1+2+3++m+ m +1 m(m + 1) m(m + 1) + 2(m + 1) (m + 1)((m + 1) + 1) = +m+1= 2 2 2 P m +1 n Sn =1+2+3++n= 6 13+23++n 3= P n =1 *« ª¸ B ( n 1+2+3++n 2 13+23++n 3= P 1 1+2+3++n 2 13=12=1 m 1+2+3++m Sn m +1 m +1 2 P m 13+23++m 3= *B( ª¸« =1+2+3++n= 3 n(n + 1) 2 2 13+23++m 3+ = = m(m + 1) 2 m+ 2 = 3 1+2+3++m + 2 m +1 3 = (m + 1)((m + 1) + 1) m 2 2 1+2+3++m+ P m +1 m +1 * ´•{0 ªÐ E · n 13+23++n 3= 8* ´ E ·ª Ð ø ¼ « [a ªÃ µ( [a 20 € 1]2 b) 2 a [0 b) 2 0 1 1+2+3++n 2 b]2 1) 2 a [0 a 0 b]2 b [a b] 1] b [a 0 1 b] [0 1) 2 1 ] 2 [0 a 1] f 1 [0 1) → 0 1 f1 x 1 2 1 = n +1 2 x 1 2 1 = n +1 2 x 1 2 1 = 4 1 n+2 2 x x=0 x= 1 2 n m n = 1mm 2 x x =1 m f2 0 1]→ 0 1 f2 x x= 1 2 n m n = 1mm 2 x m x=0 x =1 x= m 1 f 3 [0 1]→ 0 1 f3 x 2 n , n = 1,2, x 2 [0 1) 2 0 1 ] 2 [0 1] 0 1€ [a 20 1]2 b) 2 0 1 a b]2 [0 a 1 ]øª´ * ª ´* C Ó €» Z b [a b] [0 1) 2 f x = b-a x+a [a 20 11¯ = * ª ´ x = ¯ * ª ´ x -∞ g´ ¼ E · Ð * ª 1]2 b) 2 0 1 a [0 -∞ b]2 1] € +∞ 0 -∞ 0¯ = * ª ´ x a b [a b] [0 1) 2 0 +∞= * ª ´ ¯ ξ +∞ -∞ 0 f x =ex -∞ +∞ +∞ g x =-ex f ...
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## This note was uploaded on 02/06/2011 for the course CS 343 taught by Professor Zhoujunli during the Spring '08 term at BUPT.

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