The term independent of x in the expansion of

$$(1 - {x^2} + 3{x^3}){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}},\,x \ne 0$$ is :

<p>General term of Binomial expansion ${\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$ is</p> <p>$${T_{r + 1}} = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}{x^3}} \right)^{11 - r}}\,.\,{\left( { - {1 \over {5{x^2}}}} \right)^r}$$</p> <p>$$ = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}} \right)^{11 - r}}\,.\,{\left( { - {1 \over 5}} \right)^r}\,.\,{x^{33 - 5r}}$$</p> <p>In the term,</p> <p>$$\left( {1 - {x^2} + 3{x^3}} \right){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}$$</p> <p>Term independent of x is when</p> <p>(1) $33 - 5r = 0$</p> <p>$\Rightarrow r = {{33} \over 5}\,\, \notin$ integer</p> <p>(2) $33 - 5r = -2$</p> <p>$\Rightarrow 5r = 35$</p> <p>$\Rightarrow r = 7\,\, \in$ integer</p> <p>(3) $33 - 5r = - 3$</p> <p>$\Rightarrow 5r = 36$</p> <p>$\Rightarrow r = {{36} \over 5}\,\, \notin$ integer</p> <p>$\therefore$ Only for r = 7 independent of x term possible.</p> <p>$\therefore$ Independent of x term</p> <p>$$ = - \left( {{}^{11}{C_7}{{\left( {{5 \over 2}} \right)}^4}\,.\,{{\left( { - {1 \over 5}} \right)}^7}} \right)$$</p> <p>$$ = - \left( {{{11\,.\,10\,.\,9\,.\,8} \over {4\,.\,3\,.\,2\,.\,1}}\,.\,{{{5^4}} \over {{2^4}}}\,.\, - {1 \over {{5^7}}}} \right)$$</p> <p>$= {{11\,.\,10\,.\,3} \over {{2^4}\,.\,{5^3}}}$</p> <p>$= {{11\,.\,3} \over {{2^3}\,.\,{5^2}}}$</p> <p>$= {{33} \over {200}}$</p>