The coefficient of [tex]x^{-3}[/tex] in the expansion of [tex]\left(x-\frac{a}{x^2}\right)^9[/tex] is [tex]\frac{14}{9}[/tex].
(i) Find the possible values of [tex]a[/tex].
(ii) If [tex]a\ \textgreater \ 0[/tex], find the term independent of [tex]x[/tex].
(iii) Hence, for [tex]a\ \textgreater \ 0[/tex], find the term which is independent of [tex]x[/tex] in the expansion of [tex]\left(1-7 x^3\right)\left(x-\frac{a}{x^2}\right)^9[/tex].
Answer (2)
The possible values of a are 3 1 and − 3 1 . When 0"> a > 0 , the independent term found in the expansion is − 9 28 . Finally, the independent term in the combined expansion is − 14 .
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Find the general term in the binomial expansion: T r + 1 = ( r 9 ) ( − 1 ) r a r x 9 − 3 r .
Determine the possible values of a by setting the power of x to − 3 : a = ± 3 1 .
Calculate the term independent of x when a = 3 1 : Term is − 3 9 1 .
Find the term independent of x in the final expansion: The term is − 14 .
Explanation
Problem Analysis We are given the binomial expansion ( x − x 2 a ) 9 and the coefficient of x − 3 is 9 14 . We need to find the possible values of a , the term independent of x when 0"> a > 0 , and the term independent of x in the expansion of ( 1 − 7 x 3 ) ( x − x 2 a ) 9 when 0"> a > 0 .
General Term The general term in the binomial expansion of ( x − x 2 a ) 9 is given by T r + 1 = ( r 9 ) x 9 − r ( − x 2 a ) r = ( r 9 ) ( − 1 ) r a r x 9 − r − 2 r = ( r 9 ) ( − 1 ) r a r x 9 − 3 r .
Finding r for x^-3 (i) Find the possible values of a :
We want the coefficient of x − 3 , so we need 9 − 3 r = − 3 , which gives 3 r = 12 , so r = 4 .
Thus, the term with x − 3 is T 4 + 1 = T 5 = ( 4 9 ) ( − 1 ) 4 a 4 x − 3 = ( 4 9 ) a 4 x − 3 .
Calculating a^4 The coefficient of x − 3 is ( 4 9 ) a 4 = 4 ! 5 ! 9 ! a 4 = 4 × 3 × 2 × 1 9 × 8 × 7 × 6 a 4 = 126 a 4 .
We are given that the coefficient of x − 3 is 9 14 , so 126 a 4 = 9 14 .
Therefore, a 4 = 9 × 126 14 = 9 × 9 × 14 14 = 81 1 .
Finding a Taking the fourth root, we get a = ± 4 81 1 = ± 3 1 .
Finding r for the Independent Term (ii) If 0"> a > 0 , find the term independent of x :
Since 0"> a > 0 , we have a = 3 1 .
We want the term independent of x , so we need 9 − 3 r = 0 , which gives 3 r = 9 , so r = 3 .
The term independent of x is T 3 + 1 = T 4 = ( 3 9 ) ( − 1 ) 3 a 3 x 9 − 3 ( 3 ) = ( 3 9 ) ( − 1 ) 3 a 3 = 3 ! 6 ! 9 ! ( − 1 ) 3 a 3 = 3 × 2 × 1 9 × 8 × 7 ( − 1 ) 3 a 3 = 84 ( − 1 ) a 3 = − 84 a 3 .
Calculating the Independent Term Since a = 3 1 , the term independent of x is − 84 ( 3 1 ) 3 = − 84 × 27 1 = − 27 84 = − 9 28 = − 3 9 1 .
Expanding the Expression (iii) Hence, for 0"> a > 0 , find the term which is independent of x in the expansion of ( 1 − 7 x 3 ) ( x − x 2 a ) 9 :
We have ( 1 − 7 x 3 ) ( x − x 2 a ) 9 = ( x − x 2 a ) 9 − 7 x 3 ( x − x 2 a ) 9 .
Finding the Independent Term in the Second Part The term independent of x in ( x − x 2 a ) 9 is − 3 9 1 = − 9 28 .
We need to find the term independent of x in − 7 x 3 ( x − x 2 a ) 9 . This is equivalent to finding the coefficient of x − 3 in − 7 ( x − x 2 a ) 9 , and then multiplying by − 7 .
The coefficient of x − 3 in ( x − x 2 a ) 9 is 9 14 , so the term independent of x in − 7 x 3 ( x − x 2 a ) 9 is − 7 × 9 14 = − 9 98 .
Final Calculation Therefore, the term independent of x in ( 1 − 7 x 3 ) ( x − x 2 a ) 9 is − 9 28 − 9 98 = − 9 126 = − 14.
Examples
Binomial expansions are used in various fields such as physics, engineering, and computer science. For instance, in physics, they can approximate complex equations, making them easier to solve. In finance, binomial expansions can model the growth of investments over time. Understanding binomial expansions helps in simplifying and solving real-world problems across different disciplines.
