Question

The coefficient of $x^9$ in the expansion of ${\left(x^3+\frac{1}{2^{lo{g}_{\sqrt{2}}\left(x^{\frac{3}{2}}\right)}}\right)}^{11}$ is equal to

• A. $-5$
• B. $330$
• C. $520$
• D. $5+lo{g}_{\sqrt{2}}3$

Answer 1

Answer: (B)

$lo{g}_{\sqrt{2}}{x}^{\frac{3}{2}}=lo{g}_{2^{\frac{1}{2}}}{x}^{\frac{3}{2}}=\frac{\frac{3}{2}}{\frac{1}{2}}lo{g}_{2}x=lo{g}_2{x}^3$
$\Rightarrow 2^{lo{g}_{\sqrt{2}}{x}^{\frac{3}{2}}}=2^{lo{g}_2{x}^3}=x^3$
We consider the expansion of ${\left(x^3+\frac{1}{x^3}\right)}^{11}.$
$t_{r+1}{=}^{11}{C}_r{\left(x^3\right)}^{11-r}{\left(\frac{1}{x^3}\right)}^r{=}^{11}{C}_r{x}^{33-3r-3r}$
${=}^{11}{C}_r{x}^{33-6r}$
For the coefficient of $x^9,$ we get $33-6r=9$
$\Rightarrow 6r=24\Rightarrow r=4.$
Thus, the coefficient of $x^9$ is ${}^{11}{C}_4=\frac{11\times 10\times \cancel{9}3\times \cancel{8}}{\cancel{4}\times \cancel{3}\times \cancel{2}}=330.$