If $x=\large\frac{1}{3}$, the greatest term in the expansion of $(1+4x)^8$ is ?

Question

$\begin{array}{1 1} 3rd\;term \\ 4th\;term \\ 5th\;term \\ 6th\;term \end{array}$

Answer 1

General term $T_{r+1}$ in the expansion $(1+4x)^8$ = $^8C_r.4^r.x^r$

Let $(r)^{th}$ term be the greatest term.

Then $\large\frac{T_{r+1}}{T_r}$$\leq 1$

$\Rightarrow\:\large\frac{^8C_r.4^r.x^r}{^8C_{r-1}.4^{r-1}.x^{r-1}}$$\leq 1$

$\Rightarrow\:\large\frac{(9-r).4x}{r}$$\leq 1$

$\Rightarrow\:12-\large\frac{4r}{3}$$\leq r$

$\Rightarrow\:r\geq \large\frac{36}{7}$ But $r\in Z$ and $ r\leq 8$

$\Rightarrow\:r=6$

$\therefore$ The greatest term is 6th term.