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  • Help me understand! - Binomial theorem and its simple applications - JEE Main-3

If the coefficients of x^{2} and x^{3} are both zero , in the expansion of the 

expression (1+ax+bx^{2})(1-3x)^{15} in powers of x , then the 

ordered pair ( a , b ) is equal to : 

  • Option 1)

    (28, 861)

  • Option 2)

    (-54 , 315 )

  • Option 3)

    ( 28, 315 )

  • Option 4)

    ( - 21 , 714 )

 

Answers (1)

best_answer

(1+ax+bx^{2})(1-3x)^{15}=     (1+ax+bx^{2})(_{0}^{15}C\textrm{}\: -\: _{1}^{15}C\textrm{}\: 3x+_{2}^{15}C\textrm{}(3x)^{2}-..................)

coefficient of x^{2}

=> b._{0}^{15}C\textrm{}-3a \: _{1}^{15}C+9\: \: _{2}^{15}C=0

=> b - 45 a +945 = 0..........................................(1)

coefficient of x^{3}

=>-b.3._{1}^{15}C\textrm{} +9a \: _{2}^{15}C-27\: \: _{3}^{15}C=0

=> -b + 21 a -273 = 0 ........................................(2)

From (1) & (2)

a = 28 & b = 315

correct option (3).

 


Option 1)

(28, 861)

Option 2)

(-54 , 315 )

Option 3)

( 28, 315 )

Option 4)

( - 21 , 714 )

Posted by

Plabita

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