... 1, 2,K , n nên ( 1) k 1 Suy S = 2 010 ∑ ( 1) k 1 2009 k =1 k kC2 010 = 2 010 × 1) ( C k 1 2009 2009 k 1 k 1 C2009 ∀k = 1, 2,K , 2 010 i = 2 010 ∑ ( 1) C2009 = 2 010 ( − 1) i =0 i 2009 =0 0.25 ... ( 1) k C2 010 = ( 1) k k ( k ( k − 1) C k 2 010 k + kC2 010 ) với k = 2,3,K , 2 010 0.25 k k 1 Do kCn = nCn 1 ∀k = 1, 2,K , n nên ( 1) k k k k 1 k (k − 1) C2 010 = (k − 1) × 1) kC2 010 = 2 010 ... ( 1) (k − 1) C2009 ( k k = 2 010 ×2009 ( 1) k k 1 ( 1) kC2 010 = 2 010 ( 1) C2009 = −2 010 ( 1) k k k 1 k −2 k −2 C2008 ∀k = 2,3,K , 2 010 0.25 k 1 C2009 ∀k = 2,3,K , 2 010 Do 2 010 S = 2 010 ...