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  • Try this! - Binomial theorem and its simple applications - JEE Main-2

The smallest natural number n , such that the coefficient of x in

the expansion of (x^{2}+\frac{1}{x^{3}})^{n} is   ^{n}C_{23}  , is : 

  • Option 1)

    38

  • Option 2)

    58

  • Option 3)

    23

  • Option 4)

    35

 

Answers (1)

best_answer

In  the expansion of (x^{2}+\frac{1}{x^{3}})^{n}

General term is T_{r+1}=_{r}^{n}C\textrm{}(x^{2})^{(n-r)}(\frac{1}{x^{3}})^{r}

                                    =_{r}^{n}C\textrm{}(x^{2n-5r})

For coefficient of x , 2n-5r=1

                                 r=\frac{2n-1}{5}

So, we have ^{n}\textrm{C}_{\frac{2n-1}{5}} =^{n}\textrm{C}_{23}=^{n}\textrm{C}_{n-23}

  => \frac{2n-1}{5}=23 =>n=58

=>  \frac{2n-1}{5}=n-23 =>n=38

The minimum value of n be 38.

So, option (1) is correct.


Option 1)

38

Option 2)

58

Option 3)

23

Option 4)

35

Posted by

Aadil

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