Question

If $6^{th}$ term in the expansion of $(\frac{3}{2}+\frac{x}{3}{)}^n$ is the numerically greatest term when x=3, then find the sum of all possible values of n

__

Answer 1

In the binomial expansion of $(x+y{)}^n$, if we put certain values of a and b, each term will have certain value. The term which has the numerically greatest value is called the numerically greatest term. 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=$[\frac{(n+1)}{(1+|\frac{x}{y}|)}]$

$T_6$ is the numerically greatest term.
$\Rightarrow r=5$
Also r=$[\frac{(n+1)}{(1+|\frac{x}{y}|)}]$

$\frac{\text{x and y in the equation}}{\text{result is different from the one in question}}$
5=$[\frac{n+1}{(1+\frac{3\times 3}{2\times x})}]$

5=$[\frac{n+1)}{(1+\frac{3\times 3}{2\times 3})}]$

5=$[\frac{n+1}{2.5}]$

$5\le \frac{n+1}{2.5}\le 6$
$12.5\le$ n+1 < 15
$11.5\le$ n < 14
n is an integer and it can take the values 12 and 13
Sum of the possible values =12+13=25