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Find the coefficient of x in the expansion of $\left( {1 - 2{x^3} + 3{x^5}} \right){\left( {1 + \dfrac{1}{x}} \right)^8}$.

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Last updated date: 14th Jun 2024
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Answer
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Hint: Here we have to find the coefficient of x, but first of all we have to expand the given equation present in product form with x.

Complete step-by-step answer:
According to Binomial Theorem, we have
${\left( {a + b} \right)^n} = \left( {\dfrac{n}{0}} \right){a^n}{b^0} + \left( {\dfrac{n}{1}} \right){a^{n - 1}}{b^1} + .......... + \left( {\dfrac{n}{{n - 1}}} \right){a^1}{b^{n - 1}} + \left( {\dfrac{n}{n}} \right){a^0}{b^n}$
Where, in general
Let us replace a with 1, b with $\dfrac{1}{x}$ and n with 8, we get
${\left( {1 + \dfrac{1}{x}} \right)^8} = \left( {\dfrac{8}{0}} \right){1^8}\left( {\dfrac{1}{x}} \right) + \left( {\dfrac{8}{1}} \right){1^7}\left( {\dfrac{1}{x}} \right) + ......... + \left( {\dfrac{8}{7}} \right){\left( {\dfrac{1}{x}} \right)^7} + \left( {\dfrac{8}{8}} \right){\left( {\dfrac{1}{x}} \right)^8}$
$ \Rightarrow {\left( {1 + \dfrac{1}{x}} \right)^8} = \left( {\dfrac{8}{0}} \right) + \left( {\dfrac{8}{1}} \right)\left( {\dfrac{1}{x}} \right) + \left( {\dfrac{8}{2}} \right)\left( {\dfrac{1}{{{x^2}}}} \right) + \left( {\dfrac{8}{3}} \right)\left( {\dfrac{1}{{{x^3}}}} \right)......... + \left( {\dfrac{8}{7}} \right)\left( {\dfrac{1}{{{x^7}}}} \right) + \left( {\dfrac{8}{8}} \right)\left( {\dfrac{1}{{{x^8}}}} \right)$ $ \to (1)$
We have to find the coefficient of x in $\left( {1 - 2{x^3} + 3{x^5}} \right){\left( {1 + \dfrac{1}{x}} \right)^8}$
(where $\left( {1 - 2{x^3} + 3{x^5}} \right)$ is the 1st factor & ${\left( {1 + \dfrac{1}{x}} \right)^8}$ is the 2nd factor)
For coefficient of x,
‘1’ in 1st factor should multiply by a term containing x in 2nd factor but from equation(1), we can say that there is no such term
$ - 2{x^3}$ in 1st factor should multiply by a term containing \[\left( {\dfrac{1}{{{x^2}}}} \right)\] in 2nd factor, i.e. $\left( {\dfrac{8}{2}} \right)\left( {\dfrac{1}{{{x^2}}}} \right)$ term in equation(1)
$3{x^5}$ in 1st factor should multiply by a term containing \[\left( {\dfrac{1}{{{x^4}}}} \right)\] in 2nd factor, i.e. \[\left( {\dfrac{8}{4}} \right)\left( {\dfrac{1}{{{x^4}}}} \right)\] term in equation(1)
Therefore, terms containing x in the expansion $\left( {1 - 2{x^3} + 3{x^5}} \right){\left( {1 + \dfrac{1}{x}} \right)^8}$ are $\left( { - 2{x^3}} \right)\left( {\dfrac{8}{2}} \right)\left( {\dfrac{1}{{{x^2}}}} \right) \Rightarrow \left( { - 2} \right) \times \left( {\dfrac{8}{2}} \right)x$ and $\left( {3{x^5}} \right)\left( {\dfrac{8}{4}} \right)\left( {\dfrac{1}{{{x^4}}}} \right) \Rightarrow \left( 3 \right) \times \left( {\dfrac{8}{4}} \right)x$
These two terms are added in the given expansion. So the required coefficient of x can be obtained by adding the coefficients of above two terms.
By, solving $\left( {\dfrac{8}{2}} \right)$ and $\left( {\dfrac{8}{4}} \right)$ by general formula for $\left( {\dfrac{n}{r}} \right)$ , we get $\left( {\dfrac{8}{2}} \right) = 28$ and $\left( {\dfrac{8}{4}} \right) = 70$
Therefore, required coefficient of x in the expansion $ = \left( { - 2} \right) \times \left( {\dfrac{8}{2}} \right) + 3 \times \left( {\dfrac{8}{4}} \right) = \left( { - 2} \right) \times 28 + 3 \times 70 = 154$

Note: In these types of problems where coefficient of some term in the expansion is to be obtained, observe carefully what will be the terms in the factors which when multiplied will obtain the type of term asked for. Only multiply those terms to find the required coefficient. Do not solve the unnecessary terms which are not required.