Question

Numerically greatest term in the expansion of $(3x-4y)^{23}$ when $x=\frac{1}{6}$ and $y=\frac{1}{8}$ is

• A. $\frac{^{23}C_{11}}{6^{23}}$
• B. $^{23}C_{11} (\frac{8}{6})^{23}$
• C. $^{23}C_{11} (\frac{6}{8})^{23}$
• D. $^{23}C_{11}$

Hint:

For $(a+b)^n$, the numerically greatest term occurs where $|\frac{T_{r+1}}{T_r}| \le 1$. If n is odd, the two middle terms have equal maximum magnitude.

Answer 1

The Correct Option is D

Step 1: General term.
$T_{r+1} = ^{23}C_r (3x)^{23-r} (-4y)^r$

Step 2: Substitute $x=1/6, y=1/8$.
$|T_{r+1}| = ^{23}C_r (3/6)^{23-r} (4/8)^r = ^{23}C_r (1/2)^{23}$

Step 3: Maximize term magnitude.
Magnitude depends on $^{23}C_r$. Max occurs at $r = 11, 12$.

Step 4: Numerically greatest term.
$T_{12}$ or $T_{13}$, coefficient = $^{23}C_{11} = ^{23}C_{12}$