Question

If sixth term in expansion of ${(\frac{3x^2}{2}-\frac{1}{3x})}^n$ is independent of x, then the 5th term from end is:

• A. $-\frac{21}{8}$
• B. $-\frac{21x^3}{8}$
• C. $-\frac{56x^6}{4}$
• D. $-1$

Answer 1

Answer: (B)

$-\frac{21x^3}{8}$

We know the sixth term in expansion is independent

$T_{5+1}$ $=$ (coefficent)$x^{2(n-6)}$.$\frac{1}{x^6}$

$=$ (coefficent)$x^{(2n-18)}$

as it is independent of x hence $2n-18=0$ $\Rightarrow$ $n=9$

Now fifth term from end means fifth term of expansion ${(\frac{-1}{3x}+\frac{3x^2}{2})}^n$

$T_{4+1}$ $=$ ${}^9{C}_4$.${\left(\frac{-1}{3x}\right)}^{9-4}$.${\left(\frac{3x^2}{2}\right)}^4$

$=$ ${-}^9{C}_4$.$x^3$.$3^{-1}$$2^{-4}$

$=$ $-\frac{21}{8}.x^3$