If terms independent of x in the expansion of ${(3 x - \frac{1}{x})}^{20} a n d {(x + \frac{\sqrt[9]{3^{10}}}{x})}^{18}$ are A and B respectively, then $(\frac{9}{38} A + B)$ equals:

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$3^{10} .^{19} C_{8}$

$3^{9} .^{19} C_{18}$

$3^{9} .^{20} C_{8}$

$3^{10} .^{19} C_{9}$

answer is B.

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Detailed Solution

Term independent of x in expansion of ${(3 x - \frac{1}{x})}^{20}$

$T_{r + 1} =^{20} C_{r} {(3 x)}^{20 - r} {(\frac{- 1}{x})}^{r}$

When r =10, $A = T_{11} =^{20} C_{10} 3^{10} \dots . . (1)$

Term independent of x in expansion of ${(x + \frac{\sqrt[9]{3^{10}}}{x})}^{18}$

$T_{r + 1} =^{18} C_{r} {(x)}^{18 - r} {(\frac{{(3^{10})}^{1 / 9}}{x})}^{r}$

When r = 9, $B = T_{10} =^{18} C_{9} 3^{10}$

So, $(\frac{9}{37} A + B) = (\frac{9}{38} \times^{20} C_{10} \times 3^{10} +^{18} C_{9} \times 3^{10})$ [From eqns. (1) and (2)]

$= (\frac{9}{38} \times \frac{20}{10} \times \frac{19}{9} \times^{18} C_{8} \times 3^{10} +^{18} C_{9} \times 3^{10}) = 3^{10} \times^{19} C_{9}$

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