General term of $${\left[ {a{x^2} + \left( {{1 \over {bx}}} \right)} \right]^{11}}$$ is T_r+1.

T_r+1 = $${}^{11}{C_r}{\left( {a{x^2}} \right)^{11 - r}}{\left( {{1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( b \right)^{ - r}}{\left( x \right)^{22 - 3r}}$$

For the coefficient of x⁷,

$$ \Rightarrow $$ 22 - 3r = 7

$$ \Rightarrow $$ r = 5

So coefficient of x⁷ = $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ ......(1)

Now General term of $${\left[ {ax - \left( {{1 \over {b{x^2}}}} \right)} \right]^{11}}$$ is T_r+1.

T_r+1 = $${}^{11}{C_r}{\left( {a{x}} \right)^{11 - r}}{\left( { - {1 \over {bx}}} \right)^r}$$

= $${}^{11}{C_r}{\left( a \right)^{11 - r}}{\left( { - 1} \right)^r}{\left( b \right)^{ - r}}{\left( x \right)^{11 - r}}{\left( x \right)^{ - 2r}}$$

For the coefficient of x^-7,

11 - 3r = -7

$$ \Rightarrow $$ r = 6

$$\therefore$$ Coefficient of x^-7 = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$

According to question,

Coefficient of x⁷ = Coefficient of x^-7

$$ \Rightarrow $$ $${}^{11}{C_5}{\left( a \right)^6}{\left( b \right)^{ - 5}}$$ = $${}^{11}{C_6}{\left( a \right)^5}{\left( { - 1} \right)^6}{\left( b \right)^{ - 6}}$$

$$ \Rightarrow $$ $$ab = 1$$

Given $$A = \left[ {\matrix{ 1 & 0 \cr 1 & 1 \cr } } \right]$$

$$\therefore$$ $$A \times A$$ = $${A^2}$$ = $$\left[ {\matrix{ 1 & 0 \cr 2 & 1 \cr } } \right]$$

and $${A^3}$$ = $${A^2} \times A$$ = $$\left[ {\matrix{ 1 & 0 \cr 3 & 1 \cr } } \right]$$

So we can say $${A^n}$$ = $$\left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right]$$

Now $$nA - \left( {n - 1} \right){\rm I}$$

= $$\left[ {\matrix{ n & 0 \cr n & n \cr } } \right]$$ - $$\left[ {\matrix{ {n - 1} & 0 \cr 0 & {n - 1} \cr } } \right]$$

= $$\left[ {\matrix{ 1 & 0 \cr n & 1 \cr } } \right]$$ = $${A^n}$$

$$\therefore$$ $${A^n} = nA - \left( {n - 1} \right){\rm I}$$

Given, $${}^{50}{C_4} + \sum\limits_{n = 1}^6 {{}^{56 - r}{C_3}} $$

$$ \Rightarrow $$ $${}^{50}{C_4}$$ + $${}^{55}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{52}{C_3}$$ + $${}^{51}{C_3}$$ + $${}^{50}{C_3}$$

Arrange those this way

$$ \Rightarrow $$ $${}^{50}{C_4}$$ + $${}^{50}{C_3}$$ + $${}^{51}{C_3}$$ + $${}^{52}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$

We know this formula [ $${{}^n{C_r}}$$ + $${{}^n{C_{r - 1}}}$$ = $${{}^{n + 1}{C_r}}$$ ] which is used to solve this problem.

$$ \Rightarrow $$ $${}^{51}{C_4}$$ + $${}^{51}{C_3}$$ + $${}^{52}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$

$$ \Rightarrow $$ $${}^{52}{C_4}$$ + $${}^{52}{C_3}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$

$$ \Rightarrow $$$${}^{53}{C_4}$$ + $${}^{53}{C_3}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$

$$ \Rightarrow$$$$ {}^{54}{C_4}$$ + $${}^{54}{C_3}$$ + $${}^{55}{C_3}$$

$$ \Rightarrow $$$${}^{55}{C_4}$$ + $${}^{55}{C_3}$$

$$ \Rightarrow$$$$ {}^{56}{C_4}$$

$${S_n} = \sum\limits_{r = 0}^n {{1 \over {{}^n{C_r}}}}$$

=$${1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + .... + {1 \over {{}^n{C_n}}}$$

$${t_n} = \sum\limits_{r = 0}^n {{r \over {{}^n{C_r}}}}$$

= $${0 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + {2 \over {{}^n{C_2}}}.... + {n \over {{}^n{C_n}}}$$.........(1)

We can write $${t_n}$$ by rearranging like this,

$${t_n}$$ = $${n \over {{}^n{C_n}}} + {{n - 1} \over {{}^n{C_{n - 1}}}} + ... + {1 \over {{}^n{C_1}}} + {0 \over {{}^n{C_0}}}$$

= $${n \over {{}^n{C_0}}} + {{n - 1} \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {0 \over {{}^n{C_n}}}$$.........(2)

[as $${{}^n{C_0}}$$ = $${{}^n{C_n}}$$, $${{}^n{C_1}}$$ = $${{}^n{C_{n - 1}}}$$......]

By adding (1) and (2) we get,

$$2{t_n}$$ = $${n \over {{}^n{C_0}}} + {n \over {{}^n{C_1}}} + ... + {n \over {{}^n{C_{n - 1}}}} + {n \over {{}^n{C_n}}}$$

= $$n\left[ {{1 \over {{}^n{C_0}}} + {1 \over {{}^n{C_1}}} + ... + {1 \over {{}^n{C_{n - 1}}}} + {1 \over {{}^n{C_n}}}} \right]$$

= n$${S_n}$$

$$\therefore$$ $${{{t_n}} \over {{S_n}}} = {n \over 2}$$