Question

The numerically greatest term in the expansion of $(1 - 4x)^6$ at $x = \frac{3}{2}$ is

• A. 6^6
• B. 5^6
• C. $\frac{13}{2}(4)^5$
• D. $\frac{7}{2}(4)^5$

Answer 1

Answer: (A)

6^6

Answer 2

The general term in the expansion of $(1-4x)^6$ is $T_{r+1} = \binom{6}{r} (1)^{6-r} (-4x)^r$. The ratio of consecutive terms is $\frac{|T_{r+1}|}{|T_r|} = \left| \frac{6-r+1}{r} \cdot \frac{-4x}{1} \right|$. Substituting $x = \frac{3}{2}$, we get $\frac{|T_{r+1}|}{|T_r|} = \left| \frac{7-r}{r} \cdot (-6) \right| = \frac{6(7-r)}{r}$. We want to find $r$ such that $\frac{|T_{r+1}|}{|T_r|} \ge 1$, which gives $\frac{6(7-r)}{r} \ge 1$, so $42 - 6r \ge r$, leading to $42 \ge 7r$, or $6 \ge r$. This implies that $|T_1| \le |T_2| \le \dots \le |T_6| \le |T_7|$. Thus, $T_6$ and $T_7$ are the numerically greatest terms. $T_6 = \binom{6}{5} (-4)^5 (\frac{3}{2})^5 = 6 \times (-1024) \times \frac{243}{32} = -46656$. $T_7 = \binom{6}{6} (-4)^6 (\frac{3}{2})^6 = 1 \times 4096 \times \frac{729}{64} = 46656$. The numerically greatest term is $46656$, which is $6^6$.