A) 144
B) 288
C) 216
D) 576
E) \[({{2}^{11}})(3)\]
Correct Answer: D
Solution :
General term in the expansion of \[{{(1+3x+2{{x}^{2}})}^{6}}=\frac{6!}{{{r}_{1}}!{{r}_{2}}!{{r}_{3}}!}{{(1)}^{{{r}_{1}}}}\] \[{{(3x)}^{{{r}_{2}}}}{{(2{{x}^{2}})}^{{{r}_{3}}}}\] where \[{{r}_{1}}+{{r}_{2}}+{{r}_{3}}=6\] ... (i) For coefficient of\[{{x}^{11}},\]we have \[{{r}_{2}}+2{{r}_{3}}=11\] ...(ii) Now from Eqs. (i) and (ii), we get\[{{r}_{1}}={{r}_{3}}-5\] For\[{{r}_{3}}=5,{{r}_{1}}=0\]and\[{{r}_{2}}=1\] \[\therefore \]Coefficient of \[{{x}^{11}}=\frac{6!}{0!1!5!}{{(1)}^{0}}{{(3)}^{1}}{{(2)}^{5}}\] \[=6\times 3\times {{2}^{5}}=18\times 32=576\]
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