Binomial Theorem Test - 18

Total number of rational terms will be
$$1+\dfrac{6561}{L.C.M(3,9)}$$
$$=1+\dfrac{6561}{9}$$
$$=729+1$$
$$=730$$

Coefficient of $$x^7$$ in $$\displaystyle \left ( ax^2 + \frac{1}{bx} \right )^{11}$$ is $$^{11}C_5 a^6 \dfrac{1}{b^5}$$
Coefficient of $$x^{-7}$$ in $$\displaystyle \left ( ax-\frac{1}{bx^2} \right )^1$$ is $$^{11}C_6 a^5 \displaystyle \frac{1}{b^6}$$
$$^{11}C_6 a^5 \displaystyle \frac{1}{b^6}={}^{11}C_5 a^6 \displaystyle \frac{1}{b^5}$$
$$\Rightarrow ab = \displaystyle \frac{^{11}C_6}{^{11}C_5} = \frac{6}{6}=1$$

In the expansion $$(1+x)^n$$
2nd term, $$T_2=^nC_2x$$,
3rd term, $$T_3=^nC_2 \cdot x^2$$,
4th term, $$T_4=^nC_3x^3$$
Given that coefficients of these are in A.P. $$\Rightarrow 2 ^nC_2=^nC_1+^nC_3$$
$$\Rightarrow \displaystyle \frac{2n(n-1)}{2!} = n+\frac{n(n-1)(n-2)}{3!}$$
$$\Rightarrow n-1 = 1 + \displaystyle \frac{(n-1)(n-2)}{6}$$
$$\Rightarrow 6(n-1) = 6+n^2-3n+2$$
$$\Rightarrow n^2-9n+14=0     \Rightarrow n=7$$

$$T_{r+1}=\:^{n}C_{r}x^{2n-3r}$$
Therefore power of $$x$$ in the middle term is $$6$$.
Hence, $$2n-3r=6$$
$$\implies r=\dfrac{2n}{3}-2$$
The coefficient is $$924$$.
Therefore, $$\:^{n}C_{\frac{2n}{3}-2}=924$$
$$\implies \dfrac{n!}{\left(\dfrac{n}{3}+2\right)!\left(\dfrac{2n}{3}-2\right)!}=924$$
If we substitute $$n=12$$ (since it is divisible by 3),
We get $$\:^{n}C_{\frac{2n}{3}-2}$$
$$=\:^{12}C_{6}$$
$$=924$$
Hence $$n=12$$

The middle term of the expression

Coefficient of (2r + 4)th term $$=^{18}C_{2r+3}$$
Coefficient of (r - 2)th term $$=^{18}C_{r-3}$$
Coefficient of (2r + 4)th term
$$=$$ coefficient of (r - 2)th term
$$\Rightarrow ^{18}C_{2r+3}=^{18}C_{r-3}$$
$$\Rightarrow 2r+3+r-3=18$$
$$\Rightarrow 3r=18 \Rightarrow r=6$$

$${\textbf{Step  - 1: Finding middle term for both the expansion}}$$

                  $${\text{We know that, if n is even, then the middle term of (1  +  ax}}{{\text{)}}^{\text{n}}}{\text{ is given by}}{{\text{ }}^{\text{n}}}{{\text{C}}_{\frac{{\text{n}}}{{\text{2}}}}}{{\text{a}}^{\frac{{\text{n}}}{{\text{2}}}}}{{\text{x}}^{\frac{{\text{n}}}{{\text{2}}}}}$$

                  $$\therefore {\text{ Middle term of (1  + }}{\text{} \alpha}{\text{ x}}{{\text{)}}^{\text{4}}}{\text{  = }}{{\text{ }}^{\text{4}}}{{\text{C}}_{\text{2}}}{{\text{}\alpha }^{\text{2}}}{{\text{x}}^{\text{2}}}$$

                  $${\text{and Middle term of (1  +}}{\text{}  \alpha }{\text{x}}{{\text{)}}^{\text{6}}}{\text{  = }}{{\text{ }}^{\text{6}}}{{\text{C}}_{\text{3}}}{{\text{}\alpha }^{\text{3}}}{{\text{x}}^{\text{3}}}$$

$${\textbf{Step  - 2: Calculating }}{\textbf{}\alpha }$$

                  $${\text{According to question, coefficients of middle terms of both the expansion are same,}}$$$$\therefore {{\text{ }}^{\text{4}}}{{\text{C}}_{\text{2}}}{{\text{}\alpha }^{\text{2}}}{\text{  = }}{{\text{ }}^{\text{6}}}{{\text{C}}_{\text{3}}}{{\text{}\alpha }^{\text{3}}}$$

                  $$ \Rightarrow {\text{ 6}}{{\text{}\alpha }^{\text{2}}}{\text{  =  20}}{{\text{}\alpha }^{\text{3}}}$$

                  $$ \Rightarrow {\text{ }}\dfrac{{{{\text{}\alpha }^{\text{3}}}}}{{{{\text{}\alpha }^{\text{2}}}}}{\text{  =  }}\dfrac{{\text{6}}}{{{\text{20}}}}$$

                  $$ \Rightarrow {\text{ }\alpha}{\text{   =  }}\dfrac{{\text{3}}}{{{\text{10}}}}$$

$$\textbf{Hence option C is correct}$$

It can be expanded as $$\displaystyle \frac{(1-3x)^2}{1-2x}=(1-3x)^2 (1-2x)^{-1}$$
$$=(1-6x+9x^2)[1+2x+(2x)^2 + .......]$$
Therefore, coefficient of $$x^4$$ is $$2^4 - 6(2^3)+9(2^2)$$
$$=16-48+36=4$$

In the expansion $$(1+x)^n$$
5th term $$=^nC_4 \cdot x^4$$
6th term $$=^nC_5x^5$$
7th term, $$^nC_6x^6$$