標題:
Binomial
發問:
1.Let Br be the coefficient of x^r in the expansion of (1+ x)24 . If Br+2 : Br = 57 : 7 , find the value of r.2.The 7th term in the expansion of (2x+1/2x)^n in descending powers of x is a constant, where n is a positive integer. Find the value of n and the 7th term.3.Expand (1+ x)^n in ascending powers of... 顯示更多 1.Let Br be the coefficient of x^r in the expansion of (1+ x)24 . If Br+2 : Br = 57 : 7 , find the value of r. 2.The 7th term in the expansion of (2x+1/2x)^n in descending powers of x is a constant, where n is a positive integer. Find the value of n and the 7th term. 3.Expand (1+ x)^n in ascending powers of x. If three consecutive coefficients in the expansion are 3a, 13a and 39a, find the values of n and a.
最佳解答:
1. Br = C(24,24-r) = 24!/(24-r)!r! Br+2 = C(24,24-r-2) = 24!/(22-r)!(r+2)! Br+2 : Br = 57 : 7 24!/(22-r)!(r+2)! : 24!/(24-r)!r! = 57 : 7 1/(r+2)(r+1) : 1/(24-r)(23-r) = 57 : 7 57/(24-r)(23-r) = 7/(r+2)(r+1) 57/(552-47r+r2) = 7/(r2+3r+2) 3864 - 329r + 7r2 = 57r2 + 171r + 114 50r2 + 500r - 3750 = 0 r2 + 10r - 75 = 0 (r - 5)(r + 15) = 0 r = 5 or r = -15 (rejected) 2. 7th term : C(n,6) (2x)^(n-6) (1/2x)^6 = constant term x^(n - 6) (1/x)^6 = constant (n - 6) - 6 = 0 n = 12 7th term = C(12,6) (2x)^(12-6) (1/2x)^6 = 12!/6!6! = 924 3. Let the x^r term, x^(r+1) term x^(r+2) be 3a, 13a and 39a respectively. C(n,r) = 3a n!/r!(n-r)! = 3a ...... [1] C(n,r+1) = 13a n!/(r+1)!(n-r-1)! = 13a ...... [2] C(n,r+2) = 39a n!/(r+2)!(n-r-2)! = 39a ...... [3] [1]/[2] : (r+1)/(n-r) = 3/13 13r + 13 = 3n - 3r 3n = 16r + 13 ...... [4] [2]/[3] : (r+2)/(n-r-1) = 1/3 n - r - 1 = 3r + 6 n = 4r + 7 3n = 12r + 21 ...... [5] [4] = [5] : 16r + 13 = 12r + 21 4r = 8 r = 2 Put r = 2 into [4] : 3n = 16(2) + 13 n = 15 Put n = 15 and r = 2 into [1] : 15!/2!(15-2)! = 3a a = 35
其他解答:
Binomial
發問:
1.Let Br be the coefficient of x^r in the expansion of (1+ x)24 . If Br+2 : Br = 57 : 7 , find the value of r.2.The 7th term in the expansion of (2x+1/2x)^n in descending powers of x is a constant, where n is a positive integer. Find the value of n and the 7th term.3.Expand (1+ x)^n in ascending powers of... 顯示更多 1.Let Br be the coefficient of x^r in the expansion of (1+ x)24 . If Br+2 : Br = 57 : 7 , find the value of r. 2.The 7th term in the expansion of (2x+1/2x)^n in descending powers of x is a constant, where n is a positive integer. Find the value of n and the 7th term. 3.Expand (1+ x)^n in ascending powers of x. If three consecutive coefficients in the expansion are 3a, 13a and 39a, find the values of n and a.
最佳解答:
1. Br = C(24,24-r) = 24!/(24-r)!r! Br+2 = C(24,24-r-2) = 24!/(22-r)!(r+2)! Br+2 : Br = 57 : 7 24!/(22-r)!(r+2)! : 24!/(24-r)!r! = 57 : 7 1/(r+2)(r+1) : 1/(24-r)(23-r) = 57 : 7 57/(24-r)(23-r) = 7/(r+2)(r+1) 57/(552-47r+r2) = 7/(r2+3r+2) 3864 - 329r + 7r2 = 57r2 + 171r + 114 50r2 + 500r - 3750 = 0 r2 + 10r - 75 = 0 (r - 5)(r + 15) = 0 r = 5 or r = -15 (rejected) 2. 7th term : C(n,6) (2x)^(n-6) (1/2x)^6 = constant term x^(n - 6) (1/x)^6 = constant (n - 6) - 6 = 0 n = 12 7th term = C(12,6) (2x)^(12-6) (1/2x)^6 = 12!/6!6! = 924 3. Let the x^r term, x^(r+1) term x^(r+2) be 3a, 13a and 39a respectively. C(n,r) = 3a n!/r!(n-r)! = 3a ...... [1] C(n,r+1) = 13a n!/(r+1)!(n-r-1)! = 13a ...... [2] C(n,r+2) = 39a n!/(r+2)!(n-r-2)! = 39a ...... [3] [1]/[2] : (r+1)/(n-r) = 3/13 13r + 13 = 3n - 3r 3n = 16r + 13 ...... [4] [2]/[3] : (r+2)/(n-r-1) = 1/3 n - r - 1 = 3r + 6 n = 4r + 7 3n = 12r + 21 ...... [5] [4] = [5] : 16r + 13 = 12r + 21 4r = 8 r = 2 Put r = 2 into [4] : 3n = 16(2) + 13 n = 15 Put n = 15 and r = 2 into [1] : 15!/2!(15-2)! = 3a a = 35
其他解答:
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