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$\left(1-x^{2}+3 x^{3}\right)\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}, x \neq 0$ ના વિસ્તરણમાં $x$ થી સ્વતંત્ર હોય તેવું પદ.................. છે
$\frac{7}{40}$
$\frac{33}{200}$
$\frac{39}{200}$
$\frac{11}{50}$
Solution
$\left(1-x^{2}+3 x^{3}\right)\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}$
General term of $\left(\frac{5}{2} x ^{3}-\frac{1}{5 x ^{2}}\right)^{11}$ is
${ }^{11} C_{r}\left(\frac{5}{2} x^{3}\right)^{11-r}\left(-\frac{1}{5 x^{2}}\right)^{ r }$
General term is ${ }^{11} C _{ r }\left(\frac{5}{2}\right)^{11- r }\left(-\frac{1}{5}\right)^{ r } x ^{33-5 r }$
Now, term independent of $x$
$1 \times$ coefficient of $x^{0}$ in $\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}$
$-1 \times$ coefficient of $x^{-2}$ in $\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}+$
$3 \times$ coefficient of $x^{-3}$ in $\left(\frac{5}{2} x^{3}-\frac{1}{5 x^{2}}\right)^{11}$
$\begin{array}{ll}\text { for coefficient of } x ^{0} & 33-5 r =0 \text { not possible } \\ \text { for coefficient of } x ^{-2} & 33-5 r =-2 \\ & 35=5 r \Rightarrow r =7 \\ \text { for coefficient of } x ^{-3} & 33-5 r =-3 \\ & 36=5 r \text { not possible }\end{array}$
So term independent of $x$ is
$(-1)^{11} C_{7}\left(\frac{5}{2}\right)^{4}\left(-\frac{1}{5}\right)^{7}=\frac{33}{200}$
