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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