Question

The $(n+1)th$ term from the end in the expansion of ${(2x-\frac{1}{x})}^{3n}$ is

• A. $\frac{2n!}{3n!n!}\cdot 2^n\cdot x^{-n}$
• B. $\frac{n!}{2n!3n!}\cdot 2^n\cdot x^{-n}$
• C. $\frac{3n!}{2n!n!}\cdot 2^n\cdot x^{-n}$
• D. $\frac{3n!}{2n!}\cdot 2^n\cdot x^{-n}$

Answer 1

The correct option is C $\frac{3n!}{2n!n!}\cdot 2^n\cdot x^{-n}$

The expansion of the above expression will be as $T_1+T_2+....+T_{3n}+T_{3n+1}$
Now, the 1st term from the last is $T_{3n+1}$, 2nd term is $T_{3n}$,
Similarly, $(n+1)$th term from the last would be $T_{3n-(n-1)}$, i.e., $T_{2n+1}$
$T_{2n+1}{=}_{2n}^{3n}\text{C}\left(2x{)}^{3n-2n}\right((-x{)}^{-1}{)}^{2n}$
${=}_{2n}^{3n}\text{C}\left(2{)}^n\right(x{)}^n\left(x{)}^{-2n}\right(-1{)}^{2n}$
${=}_{2n}^{3n}\text{C}\left(2{)}^n\right(x{)}^{n-2n}$
$=\frac{3n!}{2n!.n!}\left(2{)}^n\right(x{)}^{-n}$