Tiêu đề 7 Binomial Theorem.pdf ✅

1.2. Use of Combination This method is the basis for deriving the binomial theorem. [DERIVATION] In expanding (a + b) n , consider each term: (a + b) n = (a + b)(a + b) ... (a + b) The a n−kb k term (where 0 ≤ k ≤ n) uses a′s (n − k of them) and b′s (k of them). The number of ways to choose is ( n k ) = n! (n − k)! k! Therefore, the expansion is (a + b) n = ∑( n k ) a n−kb k n k=0 Hence, the binomial theorem. Here is additional information based on the expansion of (a + b) n : • There are n + 1 terms in the expansion • The sum of the exponents for a and b will always be n • To get the sum of coefficients, Set all variables equal to 1 - Example: Expand (3a 2 − 2b 3 ) 6 [SOLUTION] Terms ( 6 0 ) (3a 2 ) 6 (−2b 3 ) 0 ( 6 1 ) (3a 2 ) 5 (−2b 3 ) 1 ( 6 2 ) (3a 2 ) 4 (−2b 3 ) 2 ( 6 3 ) (3a 2 ) 3 (−2b 3 ) 3 ( 6 4 ) (3a 2 ) 2 (−2b 3 ) 4 ( 6 5 ) (3a 2 ) 1 (−2b 3 ) 5 ( 6 6 ) (3a 2 ) 0 (−2b 3 ) 6 Expansion (3a 2 − 2b 3 ) 6 = 729a 12 − 2916a 10b 3 + 4860a 8b 6 − 4320a 6b 9 + 2160a 4b 12 − 576a 2b 15 + 64b 18 1.3. Use of calculator The binomial theorem applies to a calculator's "function/table" mode. - Example: Expand (3a 2 − 2b 3 ) 6 [SOLUTION] Open the calculator in the “Function/Table” Mode

(a + b) n = a n + n 1! a n−1b + n(n − 1) 2! a n−2b 2 + n(n − 1)(n − 2) 3! a n−3b 3 + n(n − 1)(n − 2)(n − 3) 4! a n−4b 4 + ⋯ - Example: (a − b 2 ) 1/3 [SOLUTION] (a − b 2 ) 1 3 = a 1 3 + 1 3 1! a − 2 3b + 1 3 (− 2 3 ) 2! a − 5 3b 2 + 1 3 (− 2 3 ) (− 5 3 ) 3! a − 8 3b 3 + 1 3 (− 2 3 ) (− 5 3 ) (− 8 3 ) 4! a − 11 3 b 4 + ⋯ (a − b 2 ) 1 3 = a 1 3 + 1 3 a − 2 3b − 1 9 a − 5 3b 2 + 5 81 a − 8 3b 3 − 10 243 a − 11 3 b 4 + ⋯ 3. Multinomial Theorem This theorem is an extension of the binomial theorem to expand the powers of multinomials. The general concept is complicated to show. Hence, this discussion shows how a term in a trinomial expansion is solved, say (x + y + z) n . [DERIVATION] (x + y + z) n = [(x + y) + z] n In any term of the expansion, n! p! (n − p)! (x + y) p z n−p From the binomial theorem, any term in (x + y) p is p! a! (p − a)! x ay p−a Substitute, n! p! (n − p)! [ p! a! (p − a)! x ay p−a ] z n−a = n! a! (n − p)! (p − a)! x ay p−a z n−a Setting b = p − a, c = n − a, any term within the expansion of (x + y + z) n is n! a! b! c! x ay b z c Where: a + b + c = n.