Question

The numerically greatest term in the expansion of $(2+3x{)}^{12}$ when $x=\frac{5}{6}$ is:

• A. ${}^{12}{C}_7{5}^5.4$
• B. ${}^{12}{C}_8\frac{5^8}{2^4}$
• C. ${}^{12}{C}_7(\frac{5}{2}{)}^7{.2}^5$
• D. None of these

Answer 1

The correct option is C

${}^{12}{C}_7(\frac{5}{2}{)}^7{.2}^5$

$(2+3x{)}^{12}atx=\frac{5}{6}=(\frac{5}{2}{)}^{12}[1+\frac{4}{5}{]}^{12}$
$|T_r|\iff |T_{r+1}|as|\frac{r}{(n-r+1)x}|\iff 1$
i.e., $|\frac{r.5}{(13-r)4}|\iff 1i.e.,r\iff \frac{52}{9}$
$r=5<\frac{52}{9}\Rightarrow T_5<T_6$
$r=6>\frac{52}{9}\Rightarrow T_6<T_7$
$\therefore$ $T_6$ is the greatest term in the expansion of $[1+\frac{4}{5}{]}^{12}$
Hence greatest term in the expansion of $(2+3x{)}^{12}$ at $x+\frac{5}{6}$ is
$(\frac{5}{12}{)}^{12}.{}^{12}{C}_5(\frac{4}{5}{)}^{5}i.e.,{}^{12}{C}_7(\frac{5}{2}{)}^7{.2}^5$