If $$n = 1,$$  then $$\frac{{\left( {n + 2} \right)!}}{{6\left( {n - 1} \right)!}} = \frac{{3!}}{{6 \times 0!}} = \frac{6}{6} = 1,$$      divisible by 1.

We observe that
\[{A^2} = \left[ {\begin{array}{*{20}{c}} 1&0\\ 2&1 \end{array}} \right],{A^3} = \left[ {\begin{array}{*{20}{c}} 1&0\\ 3&1 \end{array}} \right]\]     and we can prove by
induction that \[{A^n} = \left[ {\begin{array}{*{20}{c}} 1&0\\ n&1 \end{array}} \right]\]
Now \[nA - \left( {n - 1} \right)I = \left[ {\begin{array}{*{20}{c}} n&0\\ n&n \end{array}} \right] - \left[ {\begin{array}{*{20}{c}} {n - 1}&0\\ 0&{n - 1} \end{array}} \right]\]
\[ = \left[ {\begin{array}{*{20}{c}} 1&0\\ n&1 \end{array}} \right] = {A^n}\]
$$\therefore \,\,nA - \left( {n - 1} \right)I = {A^n}$$

$$\eqalign{ & {\text{Let, }}P\left( n \right) \equiv n! > {2^{n - 1}};P\left( 3 \right) \equiv 6 > 4 \cr & {\text{Let, }}P\left( k \right) \equiv k! > {2^{k - 1}}{\text{ is true}}{\text{.}} \cr & \therefore P\left( {k + 1} \right) = \left( {k + 1} \right)! = \left( {k + 1} \right)k! > \left( {k + 1} \right){2^{k - 1}} > {2^k}\left( {{\text{as }}k + 1 > 2} \right) \cr} $$

$$\eqalign{ & P = n\left( {n + 1} \right)\left( {n - 1} \right)\left( {n + 2} \right)\left( {n - 2} \right).....\left( {n + r} \right)\left( {n - r} \right) \cr & = \left\{ {n\left( {n + 1} \right)\left( {n + 2} \right).....\left( {n + r} \right)} \right\}\left\{ {\left( {n - 1} \right)\left( {n - 2} \right).....\left( {n - r} \right)} \right\} \cr & = \left( {n + r} \right)\left( {n + r - 1} \right).....\left( {n + 1} \right)\left( n \right)\left( {n - 1} \right).....\left( {n - r} \right) \cr} $$
Clearly $$P$$ is product of $$\left( {2r + 1} \right)$$  consecutive integers, so divisible by $$\left( {2r + 1} \right)!$$

$$\eqalign{ & {10^n} + 3\left( {{4^{n + 2}}} \right) + 5 \cr & {\text{Taking, }}n = 2;{10^2} + 3 \times {4^4} + 5 \cr & = 100 + 768 + 5 = 873 \cr} $$
Therefore this is divisible by 9.

Since, $$P\left( 5 \right)$$  is true and $$P\left( {k + 1} \right)$$   is true, whenever $$P\left( k \right)$$  is true.

For $$n = 1,P\left( 1 \right):65 + k$$     is divisible by 64.
Thus $$k,$$ should be $$ - 1$$
Since $$65 - 1 = 64$$   is divisible by 64.

$$\eqalign{ & \frac{{kn}}{{n + 1}} = \left[ {\frac{1}{{2 \times 4}} + \frac{1}{{4 \times 6}} + \frac{1}{{6 \times 8}} + .....\,n{\text{ terms}}} \right] \cr & = \frac{1}{2}\left[ {\frac{{4 - 2}}{{2 \times 4}} + \frac{{6 - 4}}{{4 \times 6}} + \frac{{8 - 6}}{{6 \times 8}} + ..... + \frac{{2n + 2 - 2n}}{{2n\left( {2n + 2} \right)}}} \right] \cr & = \frac{1}{2}\left[ {\frac{1}{2} - \frac{1}{4} + \frac{1}{4} - \frac{1}{6} + \frac{1}{6} - \frac{1}{8} + ..... + \frac{1}{{2n}} - \frac{1}{{2n + 2}}} \right] \cr & = \frac{1}{2}\left[ {\frac{1}{2} - \frac{1}{{2\left( {n + 1} \right)}}} \right] \cr & = \frac{n}{{4\left( {n + 1} \right)}} \cr & \Rightarrow k = \frac{1}{4} \cr} $$

$$\eqalign{ & {\text{Given, }}{a_0} = 1,{a_{n + 1}} = 3{n^2} + n + {a_n} \cr & \Rightarrow {a_1} = 3\left( 0 \right) + 0 + {a_0} = 1 \cr & \Rightarrow {a_2} = 3{\left( 1 \right)^2} + 1 + {a_1} = 3 + 1 + 1 = 5 \cr & {\text{From option }}\left( B \right), \cr & {\text{Let, }}P\left( n \right) = {n^3} - {n^2} + 1 \cr & \therefore P\left( 0 \right) = 0 - 0 + 1 = 1 = {a_0} \cr & P\left( 1 \right) = {1^3} - {1^2} + 1 = 1 = {a_1} \cr & {\text{and }}P\left( 2 \right) = {\left( 2 \right)^3} - {\left( 2 \right)^2} + 1 = 5 = {a_2} \cr & \therefore {a_n} = {n^3} - {n^2} + 1 \cr} $$

By taking option $$\left( d \right),$$
When $$n = 1,$$  then $$1 > \frac{1}{3}\,\,\,\left[ {{\text{true}}} \right]$$
When $$n = 2,$$  then $$5 > \frac{8}{3},\,\,\,\left[ {{\text{true}}} \right]$$
When $$n = 3,$$  then $$14 > 9,\,\,\,\left[ {{\text{true}}} \right]$$
When $$n = 4,$$  then $$30 > \frac{{64}}{3} = 21.33\,\,\,\left[ {{\text{true}}} \right]$$