The numerically greatest term in the binomial expansion of $(2 a-3 b)^{19}$ when $a=\frac{1}{4}$ and…
The numerically greatest term in the binomial expansion of $(2 a-3 b)^{19}$ when $a=\frac{1}{4}$ and $b=\frac{2}{3}$ is
- ${ }^{19} C_5 \cdot 2^{11}$
- ${ }^{19} C_3 \cdot \frac{1}{2^{11}}$
- ${ }^{19} \mathrm{C}_4 \cdot \frac{1}{2^{13}}$
- ${ }^{19} \mathrm{C}_3 \cdot 2^{13}$
Solution
We have, $(2 a-3 b)^{19}=2^{19} \cdot a^{19}\left(1-\frac{3 b}{2 a}\right)^{19}$
Ne know that, the $r^{\text {th }}$ term of greatest term of
$
\begin{aligned}
(1+x)^n & =\left[\frac{(n+1)|x|}{1+|x|}\right] \\
n & =19, x=-\frac{3 b}{2 a}
\end{aligned}
$
Here, $\quad n=19, x=-\frac{3 b}{2 a}$
$
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
$
greatest term of $(2 a-3 b)^{19}$ is
$
\begin{aligned}
& ={ }^{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} \quad\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}
\end{aligned}
$
Asked in: AP EAMCET 2018 (22 Apr Shift 1)
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