If in the expansion of

Question:

If in the expansion of $\left(x^{4}-\frac{1}{x^{3}}\right)^{15}, x^{-17}$ occurs in $r$ th term, then

(a) r = 10

(b) r = 11

(c) r = 12

(d) r = 13

Solution:

(c) r = 12

Here,

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

$=(-1)^{r} \times{ }^{15} C_{r-1} x^{64-4 r-3 r+3}$

For this term to contain $x^{-17}$, we must have :

$67-7 r=-17$

$\Rightarrow r=12$

 

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