# Find the middle term in the expansion of

Question:

Find the middle term in the expansion of $\left(\frac{\mathrm{p}}{2}+2\right)^{8}$

Solution:

Given $\mathrm{a}=\frac{\mathrm{p}}{2}$ b=2 and n=8

To find : middle term

Formula :

• The middle term $=\left(\frac{\mathrm{n}+2}{2}\right)$

$\cdot \mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

Here, n is even.

Hence,

$\left(\frac{\mathrm{n}+2}{2}\right)=\left(\frac{8+2}{2}\right)=5$

Therefore, ${ }^{5}{ }^{\text {th }}$ the term is the middle term.

For $t_{5}, r=4$

We have $\mathrm{t}_{\mathrm{r}+1}=\left(\begin{array}{l}\mathrm{n} \\ \mathrm{r}\end{array}\right) \mathrm{a}^{\mathrm{n}-\mathrm{r}} \mathrm{b}^{\mathrm{r}}$

$\therefore \mathrm{t}_{5}=\left(\begin{array}{l}8 \\ 4\end{array}\right)\left(\frac{\mathrm{p}}{2}\right)^{8-4} 2^{4}$

$\therefore \mathrm{t}_{5}=\frac{8 \times 7 \times 6 \times 5}{4 \times 3 \times 2 \times 1} \cdot\left(\frac{\mathrm{p}}{2}\right)^{4}$

$\therefore \mathrm{t}_{5}=70 \cdot\left(\frac{\mathrm{p}^{4}}{16}\right) \cdot(16)$

$\therefore \mathrm{t}_{5}=70 \mathrm{p}^{4}$

$\underline{\text { Conclusion }}$ : The middle term is $70 \mathrm{p}^{4}$.