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Q.
In the expansion of $\left(\frac{2}{x^3}+\left(a^2+4 a+6\right) x\right)^{20}, a \in R$ identify which of the following statement is(are) correct?
Binomial Theorem
Solution:
$\left(\frac{2}{x^3}+\left(a^2+4 a+6\right) x\right)^{20}$
$T _{ r +1}={ }^{20} C _{ r } \cdot\left(\frac{2}{ x ^3}\right)^{20- r } \cdot\left( a ^2+4 a +6\right)^{ r } \cdot x ^{ r }$
For term to be independent of $x$
$-60+3 r + r =0 \Rightarrow r =15 $
$\therefore T _{16}={ }^{20} C _{15} \cdot 2^5 \cdot\left( a ^2+4 a +6\right)^{15} $
$\left. T _{16}\right|_{\text {least }}={ }^{20} C _5 \cdot 2^{20}$
$ \underset{a \rightarrow-2} {\text{Lim}} \frac{\text { coefficient of } x^{20}-\text { coefficient of } x^{-60}}{(a+2)^2} $
$ \underset{a \rightarrow-2} {\text{Lim}} \frac{\left(a^2+4 a+6\right)^{20}-2^{20}}{(a+2)^2} $
$ \underset{a \rightarrow-2} {\text{Lim}} \frac{20\left(a^2+4 a+6\right)^{19} \cdot(2 a+4)}{2(a+2)}=20 \cdot 2^{19} $