Question

The middle term(s) in the expansion of ${(3x-\frac{x^3}{6})}^7$ is/are:

• A. $\frac{-105x^{13}}{8}$
• B. $\frac{35x^{15}}{48}$
• C. $\frac{35x^{13}}{73}$
• D. $\frac{105x^{15}}{48}$

Answer 1

Answer: (A), (B)

$\frac{35x^{15}}{48}$

Clearly, the given expansion has $7+1=8$ terms.
So, it has two middle terms, namely $\frac{8}{2},\frac{8}{2}+1$, i.e., $4^{\text{th}}$ and $5^{\text{th}}$ terms
Now, $T_4=T_{(3+1)}$
$=(-1{)}^3\times {}^7{C}_3\times (3x{)}^{(7-3)}\times {\left(\frac{x^3}{6}\right)}^3$
$=\frac{-105x^{13}}{8}$
And, $T_5=T_{(4+1)}$
$=(-1{)}^4\times {}^7{C}_4\times (3x{)}^{(7-4)}\times {\left(\frac{x^3}{6}\right)}^4$
$=\frac{35x^{15}}{48}$
Hence, the required middle terms are $\frac{-105x^{13}}{8}$ and $\frac{35x^{15}}{48}$