Question 331 of 347MediumMCQ
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 :
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Official solution
General term of Binomial expansion \[{\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}\] is
\[{T_{r + 1}} = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}{x^3}} \right)^{11 - r}}\,.\,{\left( { - {1 \over {5{x^2}}}} \right)^r}\]
\[ = {}^{11}{C_r}\,.\,{\left( {{5 \over 2}} \right)^{11 - r}}\,.\,{\left( { - {1 \over 5}} \right)^r}\,.\,{x^{33 - 5r}}\]
In the term,
\[\left( {1 - {x^2} + 3{x^3}} \right){\left( {{5 \over 2}{x^3} - {1 \over {5{x^2}}}} \right)^{11}}\]
Term independent of x is when
(1) 33−5r=0
\[ \Rightarrow r = {{33} \over 5}\,\, \notin \] integer
(2) 33−5r=−2
\[ \Rightarrow 5r = 35\]
\[ \Rightarrow r = 7\,\, \in \] integer
(3) 33−5r=−3
\[ \Rightarrow 5r = 36\]
\[ \Rightarrow r = {{36} \over 5}\,\, \notin \] integer
\[\therefore\] Only for r = 7 independent of x term possible.
\[\therefore\] Independent of x term
\[ = - \left( {{}^{11}{C_7}{{\left( {{5 \over 2}} \right)}^4}\,.\,{{\left( { - {1 \over 5}} \right)}^7}} \right)\]
\[ = - \left( {{{11\,.\,10\,.\,9\,.\,8} \over {4\,.\,3\,.\,2\,.\,1}}\,.\,{{{5^4}} \over {{2^4}}}\,.\, - {1 \over {{5^7}}}} \right)\]
\[ = {{11\,.\,10\,.\,3} \over {{2^4}\,.\,{5^3}}}\]
\[ = {{11\,.\,3} \over {{2^3}\,.\,{5^2}}}\]
\[ = {{33} \over {200}}\] Answer the question to unlock