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Find co-efficient of x^{5}y^{4} is the expansion of \left ( \frac{1}{2}x-2y \right )^{90}

  • Option 1)

    126

  • Option 2)

    63

  • Option 3)

    84

  • Option 4)

    None of these

 

Answers (1)

best_answer

As.learnt in concept

General Term in the expansion of (x+a)^n -

T_{r+1}= ^{n}c_{r}\cdot x^{n-r}\cdot a^{r}
 

- wherein

Where r\geqslant 0 \, and \, r\leqslant n

r= 0,1,2,----n

 

 (\frac{1}{2}x-2y)^{9}

General term = ^9C_r (\frac{1}{2}x)^{9-r}(-2y)^{r}

Here r=4

Thus we get, ^9C_4(\frac{1}{2})^5 x^{5}*(-2)^{4}y^{4}

=\frac{9!}{5!4!}*\frac{1}{2^{5}}*(-2)^{4}x^{5}y^{4}

=63x^{5}y^{4}

 


Option 1)

126

Incorrect option

Option 2)

63

Correct option

Option 3)

84

Incorrect option

Option 4)

None of these

Incorrect option

Posted by

prateek

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