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  • Need explanation for: - Binomial theorem and its simple applications - JEE Main-3

The coefficient of x^{2} in the expansion of the product

(2-x^{2})\cdot ((1+2x+3x^{2})^{6}+(1-4x^{2})^{6}) 

is

  • Option 1)

    107

  • Option 2)

    106

  • Option 3)

    108

  • Option 4)

    155

 

Answers (1)

best_answer

As we have learned

Expression of Binomial Theorem -

\left ( x+a \right )^{n}= ^{n}\! c_{0}x^{n}a^{0}+^{n}c_{1}x^{n-1}a^{1}+^{n}c_{2}x^{n-2}a^{2}x-----^{n}c_{n}x^{0}a^{n}

 

- wherein

for n  +ve integral .

\\\text{Let}\:a=\left(\left(1+2x+3x^2\right)^6+\left(1-4x^2\right)^6\right)\\\text{coefficient}\:of\:x^{2\:}\:\text{in\:the\:expension\:of}\:\\\left(2-x^2\right)\left(\left(1+2x+3x^2\right)^6+\left(1-4x^2\right)^6\right)

\\\Rightarrow 2(\text{coefficient\:of}\:x^{2\:}\text{in}\:a)-1\left(\text{constant\:of\:the\:expensaion\:in}\:a\right)\\\text{In\:the\:expension\:of} \:\left(1+2x+3x^2\right)^6+\left(1-4x^2\right)^6\:\:\text{constant}\:\:=1+1=2\\\text{Now,}\\\text{Coefficient\:of}\:x^{2\:}=\left(\text{Coefficient\:of}\:x^{2\:}\text{in}\:^6C_0\left(1+2x\right)^6\left(3x^2\right)^0\right)+\\\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left(\text{Coefficient\:of}\:x^{2\:}\text{in}\:^6C_1\left(1+2x\right)^5\left(3x^2\right)^1\right)-\left[^6C_1\left(4x^2\right)\right]\\=60+6\times 3-24=54

\\\therefore\text{ coefficient\:of}\:x^{2\:}\:\text{in\:the\:expension\:of}\:\\\left(2-x^2\right)\left(\left(1+2x+3x^2\right)^6+\left(1-4x^2\right)^6\right)\\=2\times54-2=106

 

 


Option 1)

107

This is incorrect

Option 2)

106

This is correct

Option 3)

108

This is incorrect

Option 4)

155

This is incorrect

Posted by

Himanshu

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