# Find numerically the greatest term in the expansion of

Question:

Find numerically the greatest term in the expansion of $(2+3 x)^{9}$

where . $x=\frac{3}{2}$

Solution:

To Find : numerically greatest term

For $(2+3 x)^{9}$

$a=2, b=3 x$ and $n=9$

We have relation,

$\mathrm{t}_{\mathrm{r}+1} \geq \mathrm{t}_{\mathrm{r}}$ or $\frac{\mathrm{t}_{\mathrm{r}+1}}{\mathrm{t}_{\mathrm{r}}} \geq 1$

We have a formula,

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

$=\left(\begin{array}{l}9 \\ r\end{array}\right) 2^{9-r}(3 x)^{r}$

$=\frac{9 !}{(9-\mathrm{r}) ! \times \mathrm{r} !} 2^{9-\mathrm{r}}(3)^{\mathrm{r}}(\mathrm{x})^{\mathrm{r}}$

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

$=\left(\begin{array}{c}9 \\ r-1\end{array}\right) 2^{9-r+1}(3 x)^{r-1}$

$=\frac{9 !}{(9-r+1) ! \times(r-1) !} 2^{10-r}(3)^{r-1}(x)^{r-1}$

$=\frac{9 !}{(10-r) ! \times(r-1) !} 2^{10-r}(3)^{r-1}(x)^{r-1}$

$\therefore \frac{\mathrm{t}_{\mathrm{r}+1}}{\mathrm{t}_{\mathrm{r}}} \geq 1$

$\therefore \frac{\frac{9 !}{(9-\mathrm{r}) ! \times \mathrm{r} !} 2^{9-\mathrm{r}}(3)^{\mathrm{r}}(\mathrm{x})^{\mathrm{r}}}{\frac{9 !}{(10-\mathrm{r}) ! \times(\mathrm{r}-1) !} 2^{10-\mathrm{r}}(3)^{\mathrm{r}-1}(\mathrm{x})^{\mathrm{r}-1}} \geq 1$

$\therefore \frac{9 !}{(9-r) ! \times r !} 2^{9-r}(3)^{r}(x)^{r} \geq \frac{9 !}{(10-r) ! \times(r-1) !} 2^{10-r}(3)^{r-1}(x)^{r-1}$

$\therefore \frac{9 !}{(9-r) ! \times r(r-1) !} 2^{9-r}(3)(3)^{r-1}(x)(x)^{r-1}$

$\geq \frac{9 !}{(10-\mathrm{r})(9-\mathrm{r}) ! \times(\mathrm{r}-1) !}(2) 2^{9-\mathrm{r}}(3)^{\mathrm{r}-1}(\mathrm{x})^{\mathrm{r}-1}$

$\therefore \frac{1}{r}(3)(x) \geq \frac{1}{(10-r)}(2)$

At $x=3 / 2$

$\therefore \frac{1}{\mathrm{r}}(3) \frac{3}{2} \geq \frac{1}{(10-\mathrm{r})}(2)$

$\therefore \frac{9}{4} \geq \frac{r}{(10-r)}$

$\therefore 9(10-r) \geq 4 r$

$\therefore 90-9 r \geq 4 r$

- $90 \geq 13 r$

- $r \leq 6.923$

Therefore, $r=6$ and hence the $7^{\text {th }}$ term is numerically greater.

By using formula,

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

$\mathrm{t}_{7}=\left(\begin{array}{l}9 \\ 7\end{array}\right) 2^{9-7}(3 \mathrm{x})^{7}$

$=\left(\begin{array}{l}9 \\ 2\end{array}\right) 2^{2}(3)^{7}(x)^{7}$

Conclusion : the $7^{\text {th }}$ term is numerically greater with value $\left(\begin{array}{l}9 \\ 2\end{array}\right) 2^{2}(3)^{7}(\mathrm{x})^{7}$