Question

The numerically greatest term in the binomial expansion of $(2a-3b{)}^{19}$ when $a=\frac{1}{4}$ $\mathrm{and}b=\frac{2}{3}$ is

• A. ${}^{19}{C}_5\cdot 2^{11}$
• B. ${}^{19}{C}_3\cdot \frac{1}{2^{11}}$
• C. ${}^{19}{C}_4\cdot \frac{1}{2^{13}}$
• D. ${}^{19}{C}_3\cdot 2^{13}$

Answer 1

Answer: (D)

We have,
$(2a-3b{)}^{19}=2^{19}\cdot a^{19}{\left(1-\frac{3b}{2a}\right)}^{19}$
We know that, the $r^{th}$ term of greatest term of
$(1+x{)}^n=\left[\frac{(n+1)|x|}{1+|x|}\right]$
Here, $n=19,x=-\frac{3b}{2a}$
$\therefore r=\left[\frac{(20)|\frac{3b}{2a}|}{1+\frac{3b}{2a}}\right]=\left[\frac{20}{1+4}\times 4\right]\left[\because b=\frac{2}{3},a=\frac{1}{4}\right]$
$r=16$
$\therefore$ greatest term of $(2a-3b{)}^{19}$ is
$={}^{19}{C}_{16}\times 2^{19}\cdot a^{19}{\left(\frac{3b}{2a}\right)}^{16}$
$={}^{19}{C}_3\times 2^{19}\times {\left(\frac{1}{4}\right)}^{19}\times (4{)}^{16}$
$\left[\because {}^{19}{C}_{16}={}^{19}{C}_3\right]$
$={}^{19}{C}_3\times 2^{19}\times \frac{1}{2^{38}}\times 2^{32}={}^{19}{C}_3\times 2^{13}$