Question

If terms independent of $x$ in the expansion of $\left(3 x-\frac{1}{x}\right)^{20}$ 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. $3^{10 \cdot} \cdot{ }^{19} C _8$
• B. $3^{10 \cdot{ }^{19} C _9}$
• C. $3^9 \cdot{ }^{20} C _8$
• D. $3^9 \cdot{ }^{19} C _{40}$

Answer 1

Answer: (B)

Term independent of $x$ in expansion of $\left(3 x-\frac{1}{x}\right)^{20}$
$T _{ r +1}={ }^{20} C _{ r }(3 x )^{20- r }\left(\frac{-1}{ x }\right)^{ r }$
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, $\left(\frac{9}{38} A + B \right)=\left(\frac{9}{38} \times{ }^{20} C _{10} \times 3^{10}+{ }^{18} C _9 \times 3^{10}\right)$
[From (1) and (2)]
$=\left(\frac{9}{38} \times \frac{20}{10} \times \frac{19}{9} \times{ }^{18} C _8 \times 3^{10}+{ }^{18} C _9 \times 3^{10}\right)=3^{10} \times{ }^{19} C _9 $