मराठी

Find the Coefficient Of: (Vi) X in the Expansion of ( 1 − 2 X 3 + 3 X 5 ) ( 1 + 1 X ) 8 - Mathematics

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प्रश्न

Find the coefficient of: 

(vi) x in the expansion of  \[\left( 1 - 2 x^3 + 3 x^5 \right) \left( 1 + \frac{1}{x} \right)^8\]

 
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उत्तर

 Suppose x occurs at the (+ 1)th term in the given expression.
Then, we have: 

\[(1 - 2 x^3 + 3 x^5 ) \left( 1 + \frac{1}{x} \right)^8 \]

\[ = \left( 1 - 2 x^3 + 3 x^5 \right)\left( ^{8}{}{C}_0 + ^{8}{}{C}_1 \left( \frac{1}{x} \right) + ^{8}{}{C}_2 \left( \frac{1}{x} \right)^2 + ^{8}{}{C}_3 \left( \frac{1}{x} \right)^3 +^{8}{}{C}_4 \left( \frac{1}{x} \right)^4 + ^{8}{}{C}_5 \left( \frac{1}{x} \right)^5 + ^{8}{}{C}_6 \left( \frac{1}{x} \right)^6 +^{8}{}{C}_7 \left( \frac{1}{x} \right)^7 +^{8}{}{C}_8 \left( \frac{1}{x} \right)^8 \right)\]

\[ \text{ x occurs in the above expresssion at } - 2 x^3 . ^{8}{}{C}_2 \left( \frac{1}{x^2} \right) + 3 x^5 . ^{8}{}{C}_4 \left( \frac{1}{x} \right)^4 . \]

\[ \therefore \text{ Coefficient of x }  = - 2\left( \frac{8!}{2! 6!} \right) + 3\left( \frac{8!}{4! 4!} \right) = - 56 + 210 = 154\] 

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Introduction of Binomial Theorem
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पाठ 18: Binomial Theorem - Exercise 18.2 [पृष्ठ ३७]

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आरडी शर्मा Mathematics [English] Class 11
पाठ 18 Binomial Theorem
Exercise 18.2 | Q 9.6 | पृष्ठ ३७

संबंधित प्रश्‍न

Using binomial theorem, write down the expansions  :

(iii)  \[\left( x - \frac{1}{x} \right)^6\]

\[= ^{5}{}{C}_0 (2x )^5 (3y )^0 +^{5}{}{C}_1 (2x )^4 (3y )^1 + ^{5}{}{C}_2 (2x )^3 (3y )^2 + ^{5}{}{C}_3 (2x )^2 (3y )^3 + ^{5}{}{C}_4 (2x )^1 (3y )^4 +^{5}{}{C}_5 (2x )^0 (3y )^5\]

\[= 32 x^5 + 5 \times 16 x^4 \times 3y + 10 \times 8 x^3 \times 9 y^2 + 10 \times 4 x^2 \times 27 y^3 + 5 \times 2x \times 81 y^4 + 243 y^5 \]
\[ = 32 x^5 + 240 x^4 y + 720 x^3 y^2 + 1080 x^2 y^3 + 810x y^4 + 243 y^5 \]

 

 


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