The term independent of $$x$$ in expansion of $${\left( {\frac{{x + 1}}{{{x^{\frac{2}{3}}} - {x^{\frac{1}{3}}} + 1}} - \frac{{x - 1}}{{x - {x^{\frac{1}{2}}}}}} \right)^{10}}$$ is
A.
4
B.
120
C.
210
D.
310
Answer :
210
Solution :
Given expression can be written as
$$\eqalign{
& {\left( {\left( {{x^{\frac{1}{3}}} + 1} \right) - \left( {\frac{{\sqrt x + 1}}{{\sqrt x }}} \right)} \right)^{10}} \cr
& = {\left( {{x^{\frac{1}{3}}} + 1 - 1 - \frac{1}{{\sqrt x }}} \right)^{10}} \cr
& = {\left( {{x^{\frac{1}{3}}} - {x^{ - \frac{1}{2}}}} \right)^{10}} \cr} $$
General term $$ = {T_{r + 1}} = {\,^{10}}{C_r}{\left( {{x^{\frac{1}{3}}}} \right)^{10 - r}}{\left( { - {x^{ - \frac{1}{2}}}} \right)^r}$$
$$ = {\,^{10}}{C_r}{x^{\frac{{10 - r}}{3}}}.{\left( { - 1} \right)^r}.{x^{ - \frac{r}{2}}} = {\,^{10}}{C_r}{\left( { - 1} \right)^r}.{x^{\frac{{10 - r}}{3}.\frac{r}{2}}}$$
Term will be independent of $$x$$ when $$\frac{{10 - r}}{3} - \frac{r}{2} = 0$$
$$ \Rightarrow \,\,r = 4$$
So, required term $$ = {T_5} = {\,^{10}}{C_4} = 210$$
Releted MCQ Question on Algebra >> Binomial Theorem
Releted Question 1
Given positive integers $$r > 1, n > 2$$ and that the co - efficient of $${\left( {3r} \right)^{th}}\,{\text{and }}{\left( {r + 2} \right)^{th}}$$ terms in the binomial expansion of $${\left( {1 + x} \right)^{2n}}$$ are equal. Then
If in the expansion of $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n},$$ the co-efficients of $$x$$ and $${x^2}$$ are $$3$$ and $$- 6\,$$ respectively, then $$m$$ is