Question

Find the term independent of $x$ in the expansion of $(1+x+2x^3){(\frac{3x^2}{2}-\frac{1}{3x})}^9$.

• A. $\frac{17}{4}$
• B. $\frac{1}{54}$
• C. $\frac{17}{54}$
• D. $\frac{7}{54}$

Answer 1

The correct option is C $\frac{17}{54}$

Give expression : $(1+x+2x^3){(\frac{3x^2}{2}-\frac{1}{3x})}^9$

The term independent of $x$ will have $3$ different terms

$\to 9_{C_3}{\left(\frac{3}{2}\right)}^3.x^6.{\left(\frac{-1}{3}\right)}^6{x}^{-6}+2.x^3.x^3{.9}_{C_2}{\left(\frac{3}{2}\right)}^2.x^4{\left(\frac{-1}{3}\right)}^7{x}^{-7}$

$\to 9_{C_3}\times \frac{27}{8}\times \frac{1}{3^6}+2\times 9_{C_2}\times \frac{9}{4}\times \frac{-1}{3^7}$

$\to \frac{9\times 8\times 7}{6}\times \frac{3^3}{2^2}\times \frac{1}{3^6}+2\times \frac{9\times 8}{2}\times \frac{3^2}{2^2}\times \frac{1}{3^6}+2\times \frac{9\times 8}{2}\times \frac{3^2}{2^2}\times \frac{-1}{3^7}$

$\to 7\times \frac{3^5\times 2^3}{2^4\times 3^7}-\frac{2^3\times 3^4}{2^2\times 3^7}$

$7\times \frac{1}{18}-\frac{2}{27}$

$\to \frac{21-4}{54}=\frac{17}{54}$