Question

Find the sum of coefficient of middle terms of the expansion ${(3x-\frac{x^3}{6})}^7$:

• A. $\frac{595}{48}$
• B. $-\frac{595}{48}$
• C. $-\frac{595}{24}$
• D. None of the above

Answer 1

The correct option is A $\frac{595}{48}$

Total number of terms are $8$.
So, middle term will be the $4^{th}$ and $5^{th}$ term.
$\therefore t_{3+1}{=}^7{C}_3(-1{)}^3(3x{)}^{7-3}{\left(\frac{x^3}{6}\right)}^3=-\frac{105x^{13}}{8}$
$\therefore t_{4+1}{=}^7{C}_4(-1{)}^4(3x{)}^{7-4}{\left(\frac{x^3}{6}\right)}^4=\frac{35x^{15}}{48}$
So, $-\frac{105}{8}+\frac{35}{48}=-\frac{595}{48}$
Ans-Option $A$.