The coefficient of x^{32} in the expansion of (x^{4}-\frac{1}{x^{3}})^{15}is

 

  • Option 1)

    15C_6

  • Option 2)

    15C_4

  • Option 3)

    -15C_{6}

  • Option 4)

    None

 

Answers (1)

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

 

 General term is

T_{r+1}=_{C_{r}}^{15} (x^{4})^{15-r} (\frac{-1}{x^{3}})^{r}

=_{C_{r}}^{15} x^{60-4r}\times x^{-3r}\times (-1)^{r}

=_{C{r}}^{15} x^{60-7r}\times(-1)^{r}

60-7r=32

=>r=4


Option 1)

15C_6

This is incorrect option

Option 2)

15C_4

This is correct option

Option 3)

-15C_{6}

This is incorrect option

Option 4)

None

This is incorrect option

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