Question

If terms independent of $x$ in the expansion of ${\left(3x−\frac{1}{x}\right)}^{20}$ and ${\left(x+\frac{\sqrt[9]{3^{10}}}{x}\right)}^{18}$ are A and B respectively then $\left(\frac{9}{38}A+B\right)$ equals

• A. $3^{10â‹\dots }â‹\dots {}^{19}{C}_8$
• B. $3^{10â‹\dots {}^{19}{C}_9}$
• C. $3^9â‹\dots {}^{20}{C}_8$
• D. $3^9â‹\dots {}^{19}{C}_{40}$

Answer 1

Answer: (B)

Term independent of $x$ in expansion of ${\left(3x−\frac{1}{x}\right)}^{20}$
$T_{r+1}={}^{20}{C}_r(3x{)}^{20−r}{\left(\frac{−1}{x}\right)}^r$
When $r=10$
$A=T_{11}={}^{20}{C}_{10}{3}^{10}$....(1)
Term independent of $x$ in expansion of ${\left(x+\frac{\sqrt[9]{3^{10}}}{x}\right)}^{18}$
$T_{r+1}={}^{18}{C}_r(x{)}^{18−r}{\left(\frac{{\left(3^{10}\right)}^{1/9}}{x}\right)}^r$
When $r=9$
$B=T_{10}={}^{18}{C}_9{3}^{10}$....(2)
So, $\left(\frac{9}{38}A+B\right)=\left(\frac{9}{38}×{}^{20}{C}_{10}×3^{10}+{}^{18}{C}_9×3^{10}\right)$
[From (1) and (2)]
$=\left(\frac{9}{38}×\frac{20}{10}×\frac{19}{9}×{}^{18}{C}_8×3^{10}+{}^{18}{C}_9×3^{10}\right)=3^{10}×{}^{19}{C}_9$