Find the coefficients of $x^{7}$ and $x^{8}$ in the expansion of $\left(2+\frac{x}{3}\right)^{n}$.
To find : coefficients of $x^{7}$ and $x^{8}$
Formula : $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$
Here, $a=2, b=\frac{x}{3}$
We have, $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$
$\therefore \mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right)(2)^{\mathrm{n}-\mathrm{r}}\left(\frac{\mathrm{x}}{3}\right)^{\mathrm{r}}$
$=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \frac{2^{\mathrm{n}-\mathrm{r}}}{3^{\mathrm{r}}} \mathrm{x}^{\mathrm{r}}$
To get a coefficient of $x^{7}$, we must have,
$x^{7}=x^{r}$
$\cdot r=7$
Therefore, the coefficient of $x^{7}=\left(\begin{array}{l}n \\ 7\end{array}\right) \frac{2^{n-7}}{3^{7}}$
And to get the coefficient of $x^{8}$ we must have,
$x^{8}=x^{r}$
$\cdot r=8$
Therefore, the coefficient of $x^{8}=\left(\begin{array}{l}n \\ 8\end{array}\right) \frac{2^{n-8}}{3^{8}}$
Conclusion :
- Coefficient of $x^{7}=\left(\begin{array}{l}n \\ 7\end{array}\right) \frac{2^{n-7}}{3^{7}}$
- Coefficient of $x^{8}=\left(\begin{array}{l}n \\ 8\end{array}\right) \frac{2^{n-8}}{3^{8}}$
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