1.

What is the coefficient of x17 in the expansion of \(\left(3x - \dfrac{x^3}{6}\right)^9 \ ?\)1. 189/82. 567/23. 21/164. None of the above

Answer» Correct Answer - Option 1 : 189/8

Concept:

General Term of binomial expansion = Tr+1 = nCr xn-r . yr 

 

Calculation:

Here, \(\rm \left(3x - \dfrac{x^3}{6}\right)^9 ={^9C}_r(3x)^{9-r}(-\frac{x^3}{6})^r\)

\(\rm ={^9C}_r(3^{9-r}x^{9-r})(\frac{(-x)^{3r}}{6^r})\)

\(\rm ={^9C}_r(3^{9-r}(-1)^{r}x^{9-r+3r})(\frac{1}{6^r})\)

Now we need x17 

So, 17 = 9 + 2r

⇒ 2r = 8

⇒ r = 4

Now, coefficient of x17 : \(\rm T_5={^9C}_4(3^{9-4}(-1)^{4})(\frac{1}{6^4})\)

\(\rm ={^9C}_4\times(\frac{3^{5}}{6^4})=\frac{9\times8\times7\times6\times 3}{4\times3\times2\times16}\)

=\(\frac{126\times 3}{16}\)

= 189/8

Hence, option (1) is correct.


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