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$${S_n}$$

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

Given $$\left( {1 + x} \right){\left( {1 - x} \right)^n}$$

= $${\left( {1 - x} \right)^n}$$ + $$x{\left( {1 - x} \right)^n}$$

General term of $${\left( {1 - x} \right)^n}$$ = $${}^n{C_r}.{\left( { - 1} \right)^r}.{x^r}$$

$$\therefore$$ Term containing $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ = $${}^n{C_n}.{\left( { - 1} \right)^n}.{x^n}$$

So coefficient of $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ = $${}^n{C_n}.{\left( { - 1} \right)^n}$$

General term of $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_r}.{\left( { - 1} \right)^r}.{x^{r + 1}}$$

$$\therefore$$ Term containing $${x^n}$$ in $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}.{x^n}$$

So coefficient of $${x^n}$$ in $$x{\left( {1 - x} \right)^n}$$ = $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$$

$$\therefore$$ coefficient of $${x^n}$$ in $${\left( {1 - x} \right)^n}$$ + $$x{\left( {1 - x} \right)^n}$$

= $${}^n{C_n}.{\left( { - 1} \right)^n}$$ + $${}^n{C_{n - 1}}.{\left( { - 1} \right)^{n - 1}}$$

= $${\left( { - 1} \right)^{n - 1}}\left[ {{}^n{C_{n - 1}} - {}^n{C_n}} \right]$$

= $${\left( { - 1} \right)^{n - 1}}\left[ {n - 1} \right]$$

= $${\left( { - 1} \right)^n}\left[ {1 - n} \right]$$