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If $\mathrm{x}^{4}$ occurs in the rth term in the expansion of
$\left(\mathrm{x}^{4}+\frac{1}{\mathrm{x}^{3}}\right)^{15}$, then what is the value of r?
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$\left(\mathrm{x}^{4}+\frac{1}{\mathrm{x}^{3}}\right)^{15}$, then what is the value of r?
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Verified Answer
The correct answer is:
9
In the expansion of $\left(x^{4}+\frac{1}{x^{3}}\right)^{15}$, let $T_{r}$ is the $n_{\text {th }}$
term
$\mathrm{T}_{\mathrm{r}}=15_{\mathrm{C}_{\mathrm{r}-1}}\left(\mathrm{x}^{4}\right)^{15-\mathrm{r}+1}\left(\frac{1}{\mathrm{x}^{3}}\right)^{\mathrm{r}-1}$
$=15_{\mathrm{C}_{\mathrm{r}-1}} \mathrm{x}^{64-4 \mathrm{r}-3 \mathrm{r}+3}=15_{\mathrm{C}_{\mathrm{r}-1}} \mathrm{x}^{67-7 \mathrm{r}}$
$\mathrm{x}^{4}$ occurs in this term
$\Rightarrow 4=67-7 \mathrm{r}$
$\Rightarrow 7 \mathrm{r}=63$
$\Rightarrow \quad \mathrm{r}=9$
term
$\mathrm{T}_{\mathrm{r}}=15_{\mathrm{C}_{\mathrm{r}-1}}\left(\mathrm{x}^{4}\right)^{15-\mathrm{r}+1}\left(\frac{1}{\mathrm{x}^{3}}\right)^{\mathrm{r}-1}$
$=15_{\mathrm{C}_{\mathrm{r}-1}} \mathrm{x}^{64-4 \mathrm{r}-3 \mathrm{r}+3}=15_{\mathrm{C}_{\mathrm{r}-1}} \mathrm{x}^{67-7 \mathrm{r}}$
$\mathrm{x}^{4}$ occurs in this term
$\Rightarrow 4=67-7 \mathrm{r}$
$\Rightarrow 7 \mathrm{r}=63$
$\Rightarrow \quad \mathrm{r}=9$
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