A) 10
B) 20
C) 210
D) None of these
Correct Answer: B
Solution :
The general term in the expansion of \[{{(1+2x+3{{x}^{2}})}^{10}}\] is \[\frac{10!}{r!s!t!}{{1}^{r}}{{(2x)}^{s}}{{(3{{x}^{2}})}^{t}},\] \[=\frac{10!}{r!s!t!}{{2}^{s}}\times {{3}^{t}}\times {{x}^{s+2t}}\] where \[r+s+1=10\] We have to find\[{{a}_{1}}\]ie, coefficient of\[x\]. For the coefficient of\[{{x}^{1}},\]we must have \[s+2t=1\] But \[r+s+t=10\] \[\therefore \]\[s=1-2t\]and\[r=9+t,\] where \[0\le r,s,t\le 10\] Now, \[t=0\Rightarrow s=1,r=9\] For other, values of t, we get negative value of \[s\]. So, there is only one term containing\[x\]and its coefficient is \[\frac{10!}{9!1!10!}{{2}^{1}}\times {{3}^{0}}=20\] Hence, \[{{a}_{1}}=20\]
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