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An urgent A maths question
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1a) Write down (r + 1)th term in the expansion of (√x - 1/ 3√x)^9 in descending powers of x. If we consider (a+b)^3 3C0 a^3 + 3C1 (a^2 )(b) + 3C2 (a)(b^2) + 3C3 b^3 There are 4 Terms and 3C0 is the (0+1)st Term..., 3C1 is the (1+1=2)nd Term.... So, we know The General Term of the expansion (a+b)^3 3Cr a^(3-r) * b^r Will be the r+1 term of the expansion So, the (r+1)th Term in the expansion of (√x - 1/ 3√x)^9 (Descending ) is its General Term 9Cr (√x)^(9-r) * ( - 1/ 3√x)^(r) 9Cr (√x)^(9-r) * ( - 1)^r * (3)^-r * (√x)^-r 9Cr [x^(1/2)]^(9-r) * [x^(1/2)]^-r * ( - 1)^r * (3)^-r b) Hence find the possible values of r if the power of x in this term is an integer. 9Cr (√x)^(9-r) * ( - 1)^r * (3)^-r * (√x)^-r The term involving x (√x)^(9-r) * (√x)^-r =(x^(1/2) )^(9-r) * ( x^(1/2))^-r =x^((9-r)/2) * x^(-r/2) The Power: ((9-r)/2) +(-r/2) = (9-2r )/2 Sorry... I can't find your ans 3 or 9 But if it is 3, The term involving x (√x)^(9-3) * (√x)^-3 =(x^(1/2) )^(6) * ( x^(1/2))^-3 =x^(3) * x^(-3/2) The Power : 3+ (-3/2) =3+(-1.5) =1.5 !!!??? And if it is 9, The Power will be (√x)^-9 = -4.5??? So, I am Sorry
其他解答:
(√x - 1/ 3√x)^9 = (x - 1/ 3)^9 / √x^9 (r + 1)th term = 9Cr x^(9-r) (-1/3)^r / √x^9 = 9Cr x^(9-r-9/2) (-1/3)^r = 9Cr x^(9/2-r) (-1/3)^r 9/2-r 有無可能係 integer ?? 問題有無打錯野?? 定係我理解錯??
An urgent A maths question
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Show the steps very clearly and explain it. 1a) Write down (r + 1)th term in the expansion of (√x - 1/ 3√x)^9 in descending powers of x. b) Hence find the possible values of r if the power of x in this term is an integer. Remarks: √ = square root. 更新: The question is typing correctly. Answer Key: 1b) 3 or 9 更新 2: Can anyone help me solve this question?????最佳解答:
1a) Write down (r + 1)th term in the expansion of (√x - 1/ 3√x)^9 in descending powers of x. If we consider (a+b)^3 3C0 a^3 + 3C1 (a^2 )(b) + 3C2 (a)(b^2) + 3C3 b^3 There are 4 Terms and 3C0 is the (0+1)st Term..., 3C1 is the (1+1=2)nd Term.... So, we know The General Term of the expansion (a+b)^3 3Cr a^(3-r) * b^r Will be the r+1 term of the expansion So, the (r+1)th Term in the expansion of (√x - 1/ 3√x)^9 (Descending ) is its General Term 9Cr (√x)^(9-r) * ( - 1/ 3√x)^(r) 9Cr (√x)^(9-r) * ( - 1)^r * (3)^-r * (√x)^-r 9Cr [x^(1/2)]^(9-r) * [x^(1/2)]^-r * ( - 1)^r * (3)^-r b) Hence find the possible values of r if the power of x in this term is an integer. 9Cr (√x)^(9-r) * ( - 1)^r * (3)^-r * (√x)^-r The term involving x (√x)^(9-r) * (√x)^-r =(x^(1/2) )^(9-r) * ( x^(1/2))^-r =x^((9-r)/2) * x^(-r/2) The Power: ((9-r)/2) +(-r/2) = (9-2r )/2 Sorry... I can't find your ans 3 or 9 But if it is 3, The term involving x (√x)^(9-3) * (√x)^-3 =(x^(1/2) )^(6) * ( x^(1/2))^-3 =x^(3) * x^(-3/2) The Power : 3+ (-3/2) =3+(-1.5) =1.5 !!!??? And if it is 9, The Power will be (√x)^-9 = -4.5??? So, I am Sorry
其他解答:
(√x - 1/ 3√x)^9 = (x - 1/ 3)^9 / √x^9 (r + 1)th term = 9Cr x^(9-r) (-1/3)^r / √x^9 = 9Cr x^(9-r-9/2) (-1/3)^r = 9Cr x^(9/2-r) (-1/3)^r 9/2-r 有無可能係 integer ?? 問題有無打錯野?? 定係我理解錯??
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