Question

The numerically greatest term in the binomial expansion of $(2 a-3 b)^{19}$ when $a=\frac{1}{4}$ $\operatorname{and} b=\frac{2}{3}$ 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,
$(2 a-3 b)^{19}=2^{19} \cdot a^{19}\left(1-\frac{3 b}{2 a}\right)^{19}$
We know that, the $r^{th}$ term of greatest term of
$( 1 +x)^{n}=\left[\frac{(n+ 1 )|x|}{ 1 +|x|}\right]$
Here, $n=19, \,x=-\frac{3 b}{2 a}$
$\therefore \,r=\left[\frac{(20)\left|\frac{3 b}{2 a}\right|}{1+\frac{3 b}{2 a}}\right]=\left[\frac{20}{1+4} \times 4\right] \quad\left[\because b=\frac{2}{3}, a=\frac{1}{4}\right]$
$r=16$
$\therefore $ greatest term of $(2 a-3 b)^{19}$ is
$={ }^{19} C_{16} \times 2^{19} \cdot a^{19}\left(\frac{3 b}{2 a}\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} $