Find the coefficients of

Question:

Find the coefficients of $x^{7}$ and $x^{8}$ in the expansion of $\left(2+\frac{x}{3}\right)^{n}$.

Solution:

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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