Find the coefficient of the terms x^-3 and x in the binomial (3x-a/x)^5 using the formula and without expanding?
回答 (1)
2017-08-23 10:33 pm
The general term in expansion of [3x - (a/x)]^5, T(r+1)
= C(5,r) * (3x)^(5-r) * (a/x)^r
= [C(5,r) * 3^(5-r) * a^r] * x^(5-2r)
For the x^(-3) term :
x^(5 - 2r) = x^(-3)
5 - 2r = -3
r = 4
Coefficient of x^(-3) term
= C(5,4) * 3^(5-4) * a^4
= 15 a^4
For the x term :
x^(5 - 2r) = x
5 - 2r = 1
r = 2
Coefficient of x term
= C(5,2) * 3^(5-2) * a^2
= 270 a^2
= C(5,r) * (3x)^(5-r) * (a/x)^r
= [C(5,r) * 3^(5-r) * a^r] * x^(5-2r)
For the x^(-3) term :
x^(5 - 2r) = x^(-3)
5 - 2r = -3
r = 4
Coefficient of x^(-3) term
= C(5,4) * 3^(5-4) * a^4
= 15 a^4
For the x term :
x^(5 - 2r) = x
5 - 2r = 1
r = 2
Coefficient of x term
= C(5,2) * 3^(5-2) * a^2
= 270 a^2
2017-08-23 9:57 pm
(k+1)th term=(5Ck) *[(3x) ^(5-k) ]*[(-a/x)^k]
=(5Ck)*[3^(5-k) ]*[(-a) ^k]*[x^(5-2k) ]
x^(-3) --->5-2k= -3 --->k=4
--->its coefficient =(5C4)*3*(a^4)
= 15a^4
=(5Ck)*[3^(5-k) ]*[(-a) ^k]*[x^(5-2k) ]
x^(-3) --->5-2k= -3 --->k=4
--->its coefficient =(5C4)*3*(a^4)
= 15a^4
收錄日期: 2021-04-18 17:43:46
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https://hk.answers.yahoo.com/question/index?qid=20170823131023AAR1D0N