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  • Solve it, - Binomial theorem and its simple applications - JEE Main

If the number of terms in the expansion of

\left ( 1-\frac{2}{x} +\frac{4}{x^{2}}\right )^{n},\: x\neq 0

28, then the sum of the coefficients of all the terms in this expansion, is

  • Option 1)

    64

  • Option 2)

     2187

     

  • Option 3)

     243

     

  • Option 4)

     729

     

 

Answers (2)

best_answer

As we learnt in

Sum of Binomial Coefficients -

(x+a)^{n}= ^{n}c_{0}x^{n}a^{0}+^{n}c_{r}x^{n-1}a+^{n}c_{2}x^{n-2}a^{2}+---

x= a= 1:

\therefore c_{0}+c_{1}+c_{2}+c_{3}+----= 2^{n}

-

 

 ln \left(\frac{1-2}{x}+\frac{4}{x^{2}} \right )^{n}

Number of terms ^{n+2}C_{2}=28

Thus n = 6

Also, in \left(\frac{1-2}{x}+\frac{4}{x^{2}} \right )^{n}

Put x = 1, x = 6

We get sum of coefficients =3^{6}=729

Correct option is 4.

 


Option 1)

64

This is an incorrect option.

Option 2)

 2187

 

This is an incorrect option.

Option 3)

 243

 

This is an incorrect option.

Option 4)

 729

 

This is the correct option.

Posted by

Aadil

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64

Posted by

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