$${T_{r + 1}}$$  in the expansion
$$\eqalign{ & {\left[ {a{x^2} + \left( {\frac{1}{{bx}}} \right)} \right]^{11}} = {\,^{11}}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {\frac{1}{{bx}}} \right)^r} \cr & = {\,^{11}}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 2r - r}} \cr} $$
For the Co-efficient of $${x^7},$$ we have
$$\eqalign{ & \Rightarrow \,\,22 - 3r = 7 \cr & \Rightarrow \,\,r = 5 \cr} $$
$$\therefore \,\,{\text{Co-eff}}{\text{. of }}{x^7} = {\,^{11}}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}\,\,\,\,\,\,\,\,.....\left( 1 \right)$$
Again $${T_{r + 1}}$$  in the expansion
$$\eqalign{ & {\left[ {ax - \frac{1}{{b{x^2}}}} \right]^{11}} = {\,^{11}}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( { - \frac{1}{{b{x^2}}}} \right)^r} \cr & = {\,^{11}}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r} \times {\left( b \right)^{ - r}}{\left( x \right)^{ - 2r}}{\left( x \right)^{11 - r}} \cr} $$
For the Co-efficient of $${x^{ - 7}},$$ we have
$$\eqalign{ & {\text{Now }}11 - 3r = - 7 \cr & \Rightarrow \,\,3r = 18 \cr & \Rightarrow \,\,r = 6 \cr & \therefore \,\,{\text{Co-eff}}{\text{. of }}{x^{ - 7}} = {\,^{11}}{C_6}{a^5} \times 1 \times {\left( b \right)^{ - 6}} \cr & \therefore \,\,{\text{Co-eff}}{\text{. of }}{x^7} = {\text{Co-eff}}{\text{. of }}{x^{ - 7}} \cr & \Rightarrow \,{\,^{11}}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}} \cr & = \,{\,^{11}}{C_6}{a^5} \times {\left( b \right)^{ - 6}} \cr & \Rightarrow \,\,ab = 1. \cr} $$

Let $$r + 1 = 7$$
$$ \Rightarrow r = 6$$
Given expansion is $${\left( {\frac{3}{{\root 3 \of {84} }} + \sqrt 3 \ln x} \right)^9},x > 0$$
We have $${T_{r + 1}} = {\,^n}{C_r}{\left( x \right)^{n - r}}{a^r}{\text{ for }}{\left( {x + a} \right)^n}.$$
$$\therefore $$ According to the question
$$\eqalign{ & 729 = {\,^9}{C_6}{\left( {\frac{3}{{\root 3 \of {84} }}} \right)^3} \cdot {\left( {\sqrt 3 \ln x} \right)^6} \cr & \Rightarrow {3^6} = 84 \times \frac{{{3^3}}}{{84}} \times {3^3} \times \left( {6\ln x} \right) \cr & \Rightarrow {\left( {\ln x} \right)^6} = 1 \cr & \Rightarrow {\left( {\ln x} \right)^6} = {\left( {\ln e} \right)^6} \cr & \Rightarrow x = e \cr} $$

$$\eqalign{ & {\text{Clearly }}{A_r} = {\,^{10}}{C_r},{B_r} = {\,^{20}}\,{C_r},{C_r} = {\,^{30}}{C_r} \cr & {\text{Now }}\sum\limits_{r = 1}^{10} {^{10}{C_r}\left( {^{20}{C_{10}}{\,^{20}}{C_r} - {\,^{30}}{C_{10}}^{10}{C_r}} \right)} \cr & { = ^{20}}{C_{10}}\sum\limits_{r = 1}^{10} {^{10}{C_r}{\,^{20}}{C_r}\, - {\,^{30}}{C_{10}}\sum\limits_{r = 1}^{10} {^{10}{C_r}{ \times ^{10}}{C_r}} } \cr & = {\,^{20}}{C_{10}}\left( {^{10}{C_1}{\,^{20}}{C_1} + {\,^{10}}{C_2}{\,^{20}}{C_2} + .....{ + ^{10}}{C_{10}}{\,^{20}}{C_{10}}} \right) - {\,^{30}}{C_{10}}\left( {^{10}{C_1} \times {\,^{10}}{C_1} + {\,^{10}}{C_2} \times {\,^{20}}{C_2} + ..... + {\,^{10}}{C_{10}} \times {\,^{10}}{C_{10}}} \right)\,\,\,\,.....\left( 1 \right) \cr} $$
Now expanding $${\left( {1 + x} \right)^{10}}\,{\text{and }}{\left( {1 + x} \right)^{20}}$$     by binomial theorem and comparing the coefficients of $${x^{20}}$$ in their product, on both sides, we get
$$\eqalign{ & ^{10}{C_0}{\,^{20}}{C_0} + {\,^{10}}{C_1}{\,^{20}}{C_1} + {\,^{10}}{C_2}{\,^{20}}{C_2} + ..... + {\,^{10}}{C_{10}}{\,^{20}}{C_{10}} \cr & = \,{\text{co - eff}}{\text{. of }}{x^{20}}\,{\text{in }}{\left( {1 + x} \right)^{30}} = {\,^{30}}{C_{20}} = {\,^{30}}{C_{10}} \cr & \therefore \,{\,^{10}}{C_1}{\,^{20}}{C_1} + {\,^{10}}{C_2}{\,^{20}}{C_2} + ..... + {\,^{10}}{C_{10}}{\,^{20}}{C_{10}} = {\,^{30}}{C_{10}} - 1 \cr} $$
Again expending $${\left( {1 + x} \right)^{10}}\,{\text{and }}{\left( {x + 1} \right)^{10}}$$     by binomial theorem and comparing the coefficients of $${x^{10}}$$ in their product on both sides, we get
$$\eqalign{ & \therefore \,\,{\left( {^{10}{C_0}} \right)^2}{\left( {^{10}{C_1}} \right)^2} + {\left( {^{10}{C_2}} \right)^2} + ..... + {\left( {^{10}{C_{10}}} \right)^2} = \,{\text{co - eff}}{\text{. of }}{x^{10}}\,{\text{in}}\,{\left( {1 + x} \right)^{20}} = {\,^{20}}{C_{10}} \cr & \therefore \,\,{\left( {^{10}{C_1}} \right)^2} + {\left( {^{10}{C_2}} \right)^2} + ..... + {\left( {^{10}{C_{10}}} \right)^2} = {\,^{20}}{C_{10}} - 1 \cr} $$
Substituting these values in equation (1), we get
$$\eqalign{ & = {\,^{20}}{C_{10}}\left( {^{30}{C_{10}} - 1} \right) - {\,^{30}}{C_{10}}\left( {^{20}{C_{10}} - 1} \right) \cr & = {\,^{30}}{C_{10}} - {\,^{20}}{C_{10}} = \,{C_{10}} - {B_{10}} \cr} $$

To find
$$^{30}{C_0}^{30}{C_{10}} - {\,^{30}}{C_1}^{30}{C_{11}} + {\,^{30}}{C_2}^{30}{C_{12}} - ..... + {\,^{30}}{C_{20}}^{30}{C_{30}}$$
We know that
$$\eqalign{ & {\left( {1 + x} \right)^{30}} = {\,^{30}}{C_0} + {\,^{30}}{C_1}x + {\,^{30}}{C_2}{x^2} + ..... + {\,^{30}}{C_{20}}{x^{20}} + .....{\,^{30}}{C_{30}}{x^{30}}\,\,\,\,.....\left( 1 \right) \cr & {\left( {x - 1} \right)^{30}} = {\,^{30}}{C_0}{x^{30}} - {\,^{30}}{C_1}{x^{29}} + ..... + {\,^{30}}{C_{10}}{x^{20}} - {\,^{30}}{C_{11}}{x^{19}} + {\,^{30}}{C_{12}}{x^{18}} + .....{\,^{30}}{C_{30}}{x^0}\,\,\,.....\left( 2 \right) \cr} $$
Multiplying eqn. (1) and (2), we get
$${\left( {{x^2} - 1} \right)^{30}} = \left( 1 \right) \times \left( 2 \right)$$
Equating the coefficients of $$x^{20}$$ on both sides, we get
$$\eqalign{ & ^{30}{C_{10}} = {\,^{30}}{C_0}^{30}{C_{10}} - {\,^{30}}{C_1}^{30}{C_{11}} + {\,^{30}}{C_2}^{30}{C_{12}} - ..... + {\,^{30}}{C_{20}}^{30}{C_{30}} \cr & \therefore {\text{Req}}{\text{. value is}}{{\text{ }}^{30}}{C_{10}} \cr} $$

$$\eqalign{ & \frac{2}{1} \cdot \frac{1}{3} + \frac{3}{2} \cdot \frac{1}{9} + \frac{4}{3} \cdot \frac{1}{{27}} + \frac{5}{4} \cdot \frac{1}{{81}} + .....\infty \cr & = \left( {1 + 1} \right)\frac{1}{3} + \left( {1 + \frac{1}{2}} \right){\left( {\frac{1}{3}} \right)^2} + \left( {1 + \frac{1}{3}} \right){\left( {\frac{1}{3}} \right)^3} + \left( {1 + \frac{1}{4}} \right){\left( {\frac{1}{3}} \right)^4} + .....\infty \cr & = \left\{ {\frac{1}{3} + {{\left( {\frac{1}{3}} \right)}^2} + {{\left( {\frac{1}{3}} \right)}^3} + .....\infty } \right\} + \left\{ {\left( {\frac{1}{3}} \right) + \frac{1}{2}{{\left( {\frac{1}{3}} \right)}^2} + \frac{1}{3}{{\left( {\frac{1}{3}} \right)}^3} + .....\infty } \right\} \cr & = \frac{{\frac{1}{3}}}{{1 - \frac{1}{3}}} - {\log _e}\left( {1 - \frac{1}{3}} \right) = \frac{1}{2} - {\log _e}\left( {\frac{2}{3}} \right) \cr & = \frac{1}{2} + {\log _e}3 - {\log _e}2 \cr} $$

Expression $$ = {10^n} - 1 + 3 \cdot {4^{n + 2}} + 6 = \left( {{{10}^n} - 1} \right) + 6\left( {{2^{2n + 3}} + 1} \right)$$
$$\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, = \left( {10 - 1} \right)\left\{ {{{10}^{n - 1}} + {{10}^{n - 2}} + ..... + 10 + 1} \right\} + 6\left( {2 + 1} \right)\left\{ {{2^{2n + 2}} - {2^{2n + 1}} + ..... - 2 + 1} \right\}.$$
$${10^{n - 1}} + {10^{n - 2}} + ..... + 10 + 1$$      is an odd number which is not divisible by any odd number for all $$n \cdot {2^{2n + 2}} - {2^{2n + 1}} + ..... - 2 + 1$$      is also an odd number which is not divisible by any odd number for all $$n.$$
∴ the expression is divisible by $$9.$$

We have $$a =$$  sum of the coefficient in the expansion $${\left( {1 - 3x + 10{x^2}} \right)^n} = {\left( {1 - 3 + 10} \right)^n} = {\left( 8 \right)^n}$$
$$ \Rightarrow {\left( {1 - 3x + 10{x^2}} \right)^n} = {\left( 2 \right)^{3n}},\left[ {{\text{Putting }}x = 1} \right]$$
Now, $$b =$$  sum of the coefficient in the expansion of $${\left( {1 + {x^2}} \right)^n} = {\left( {1 + 1} \right)^n} = 2n.\,\,{\text{Clearly, }}a = {b^3}$$

We have $${\left( {1 + x} \right)^m}{\left( {1 - x} \right)^n}$$
$$\eqalign{ & \left[ {1 + mx + \frac{{m\left( {m - 1} \right)}}{{2!}}{x^2} + .....} \right]\left[ {1 - nx + \frac{{n\left( {n - 1} \right)}}{{2!}}{x^2} - .....} \right] \cr & = 1 + \left( {m - n} \right)x + \left[ {\frac{{m\left( {m - 1} \right)}}{2} + \frac{{n\left( {n - 1} \right)}}{2} - mn} \right]{x^2} + ..... \cr & {\text{Given, }}m - n = 3\,\,\,\,\,\,\,\,.....\left( 1 \right) \cr & {\text{and }}\frac{1}{2}m\left( {m - 1} \right) + \frac{1}{2}n\left( {n - 1} \right) - mn = - 6 \cr & \Rightarrow \,\,{m^2} + {n^2} - 2mn - \left( {m + n} \right) = - 12 \cr & \Rightarrow \,\,{\left( {m - n} \right)^2} - \left( {m + n} \right) = - 12 \cr & \Rightarrow \,\,m + n = 9 + 12 = 21\,\,\,\,\,\,\,\,.....\left( 2 \right) \cr} $$
From (1) and (2), we get
$$m = 12$$

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 $${x^4} = {\,^{10}}{C_2}\frac{{{{\left( { - 1} \right)}^2}{3^2}}}{{{2^8}}} = \frac{{405}}{{256}}$$

$$\eqalign{ & \left( {^{10}{C_0}} \right) + \left( {^{10}{C_0} + {\,^{10}}{C_1}} \right) + \left( {^{10}{C_0} + {\,^{10}}{C_1} + {\,^{10}}{C_2}} \right) + ..... + \left( {^{10}{C_0} + {\,^{10}}{C_1} + {\,^{10}}{C_2} + ..... + {\,^{10}}{C_9}} \right) \cr & = \,10 {\,^{10}}{C_0} + 9 {\,^{10}}{C_1} + 8 {\,^{10}}{C_2} + ..... + {\,^{10}}{C_9} \cr & = {\,^{10}}{C_1} + 2{\,^{10}}{C_2} + 3{\,^{10}}{C_3} + ..... + 10{\,^{10}}{C_{10}} \cr & = \,\sum\limits_{r = 1}^{10} {r{\,^{10}}} {C_r} = 10\sum\limits_{r = 1}^{10} {^9} {C_{r - 1}} = 10 \cdot {2^9} \cr} $$