Question

In the expansion ${\left(1-3x+3x^2-x^3\right)}^{2n}$, the middle term is

• A. ${(n+1)}^{th}term$
• B. ${(2n+1)}^{th}term$
• C. ${(3n+1)}^{th}term$
• D. None of these

Answer 1

The correct option is C

${(3n+1)}^{th}term$

Explanation For The Correct Option:

Finding the middle term In the expansion ${\left(1-3x+3x^2-x^3\right)}^{2n}$:

$\Rightarrow$$$\begin{gathered} \Rightarrow {\left(1-3x+3x^2-x^3\right)}^{2n}={\left[1^3-x^3-3x(1-x)\right]}^{2n} \\ ={\left[{\left(1-x\right)}^3\right]}^{2n}[\therefore {(a-b)}^3=a^3-b^3-3ab(a-b\left)\right] \\ ={\left(1-x\right)}^{6n} \end{gathered}$$

$6n$is an even number then $(6n+1)$will be an odd number. So the expansion will have odd number of terms.

In this case, the middle term will be the ${\left(\frac{6n}{2}+1\right)}^{th}={\left(3n+1\right)}^{th}$ term.

Hence, option C is the correct option.