The number of terms whose values depend on $$x$$ in the expansion of $${\left( {{x^2} - 2 + \frac{1}{{{x^2}}}} \right)^n}$$ is
A.
$$2n + 1$$
B.
$$2n$$
C.
$$n$$
D.
None of these
Answer :
$$2n$$
Solution :
$$\eqalign{
& {\left( {{x^2} - 2 + \frac{1}{{{x^2}}}} \right)^n} = {\left( {x - \frac{1}{x}} \right)^{2n}} \cr
& \therefore \,\,{t_{r + 1}} = {\,^{2n}}{C_r}{x^{2n - r}} \cdot {\left( { - \frac{1}{x}} \right)^r} = {\left( { - 1} \right)^r} \cdot {\,^{2n}}{C_r} \cdot {x^{2n - 2r}}. \cr} $$
If $$2n - 2r = 0,$$ the value of the term will not depend on $$x.$$ So, $$r = n.$$
∴ the required number of terms $$ = \left( {2n + 1} \right) - 1 = 2n.$$
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