Question

Find the numerically greatest term in the expansion of $(4+6x{)}^{24}$ when $x=\frac{1}{6}$.

• A.
• B.
• C.
• D.

Answer 1

Answer: (B), (C)

The correct options are
B

C

Consider the expansion of $(x+y{)}^n$. Let us assume $T_{r+1}$ has the numerically greatest term. Then, $|T_{r+1}|\ge |T_r|$

$\Rightarrow |\frac{T_{r+1}}{T_r}|\ge 1$
$\Rightarrow |\frac{{}^n{C}_r{x}^{n-r}{y}^r}{{}^n{C}_{r-1}{x}^{n-r+1}{y}^{r-1}}|\ge 1$

$\Rightarrow \frac{n-r+1}{r}|\frac{y}{x}|\ge 1$

$\Rightarrow$ When we solve for r, we get
r=$[\frac{(n+1)}{(1+|\frac{x}{y}|)}]$. The greatest term occurs for r=$[\frac{(n+1)}{(1+|\frac{x}{y}|)}]$, where [] denotes the greatest integer fuction. If $\frac{(n+1)}{(1+|\frac{x}{y}|)}$ is an integer, then $T_r$ and $T_{r+1}$ both are greatest terms.

$\frac{(n+1)}{(1+|\frac{x}{y}|)}=[\frac{\left(12\right)}{\left(3.1\right)}]$ (x on the LHS and RHS are different)
When $x=\frac{1}{6},6x=1$
$\Rightarrow \frac{(n+1)}{(1+|\frac{x}{y}|)}=\frac{\left(25\right)}{(1+4)})$
=5
$\frac{(n+1)}{(1+|\frac{x}{y}|)}$ is an integer $\Rightarrow T_{r}and{T}_{r+1}$ both are greatest terms.

Or $T_{5}and{T}_6$ both are greatest terms.