Question

The co-efficient of $${x^4}$$ in $${\left( {\frac{x}{2} - \frac{3}{{{x^2}}}} \right)^{10}}$$   is

A. $$\frac{{405}}{{256}}$$  
B. $$\frac{{504}}{{259}}$$
C. $$\frac{{450}}{{263}}$$
D. none of these
Answer :   $$\frac{{405}}{{256}}$$
Solution :
General term in the expansion $${\left( {\frac{x}{2} - \frac{3}{{{x^2}}}} \right)^{10}}$$   is
$${T_{r + 1}} = {\,^{10}}{C_r}{\left( {\frac{x}{2}} \right)^{10 - r}}{\left( {\frac{{ - 3}}{{{x^2}}}} \right)^r} = {\,^{10}}{C_r}{x^{10 - 3r}}\frac{{{{\left( { - 1} \right)}^r}{3^r}}}{{{2^{10 - r}}}}$$
For co - eff of $${x^4},$$ we should have
$$\eqalign{ & 10 - 3r = 4 \cr & \Rightarrow \,\,r = 2 \cr} $$
∴ Co - eff of $${x^4} = {\,^{10}}{C_2}\frac{{{{\left( { - 1} \right)}^2}{3^2}}}{{{2^8}}} = \frac{{405}}{{256}}$$

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

A. $$n = 2r$$
B. $$n = 2r + 1$$
C. $$n = 3r$$
D. none of these
Releted Question 2

The co-efficient of $${x^4}$$ in $${\left( {\frac{x}{2} - \frac{3}{{{x^2}}}} \right)^{10}}$$   is

A. $$\frac{{405}}{{256}}$$
B. $$\frac{{504}}{{259}}$$
C. $$\frac{{450}}{{263}}$$
D. none of these
Releted Question 3

The expression $${\left( {x + {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5} + {\left( {x - {{\left( {{x^3} - 1} \right)}^{\frac{1}{2}}}} \right)^5}$$       is a polynomial of degree

A. 5
B. 6
C. 7
D. 8
Releted Question 4

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

A. 6
B. 9
C. 12
D. 24

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