It can be proved with the help of mathematical induction that $$\frac{n}{2} > a\left( n \right) \leqslant n.$$
$$\eqalign{ & \therefore \frac{{200}}{2} < a\left( {200} \right) \cr & \Rightarrow a\left( {200} \right) > 100{\text{ and }}a\left( {100} \right) \leqslant 100. \cr} $$

When $$n = 2$$  then $${\left( {\frac{{n + 1}}{2}} \right)^n} = \frac{9}{4}$$
$$ \Rightarrow n! < {\left( {\frac{{n + 1}}{2}} \right)^n}$$
When $$n = 3,$$  then $$n! = 6,{\left( {\frac{{n + 1}}{2}} \right)^n} = 8$$
$$ \Rightarrow n! < {\left( {\frac{{n + 1}}{2}} \right)^n}$$
When $$n = 4,$$  then $$n! = 24,$$
$$\eqalign{ & {\left( {\frac{{n + 1}}{2}} \right)^n} = \frac{{625}}{{16}} \cr & \Rightarrow n! < {\left( {\frac{{n + 1}}{2}} \right)^n} \cr} $$
$$\therefore $$ It is seen that $$n! < {\left( {\frac{{n + 1}}{2}} \right)^n}$$

Let the statement $$P\left( n \right)$$  be defined as
$$\eqalign{ & P\left( n \right):1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + ..... + \frac{1}{{1 + 2 + 3 + ..... + n}} = \frac{{2n}}{{n + 1}} \cr & {\text{i}}{\text{.e}}{\text{., }}P\left( n \right):1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + ..... + \frac{2}{{n\left( {n + 1} \right)}} = \frac{{2n}}{{n + 1}} \cr} $$
Step I : For $$n = 1,P\left( 1 \right):1 = \frac{{2 \times 1}}{{1 + 1}} = \frac{2}{2} = 1,$$
which is true.
Step II : Let it is true for $$n = k,$$  i.e., $$1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + ..... + \frac{2}{{k\left( {k + 1} \right)}} = \frac{{2k}}{{k + 1}}\,\,\,.....\left( {\text{i}} \right)$$
Step III : For $$n = k + 1,\left( {1 + \frac{1}{{1 + 2}} + \frac{1}{{1 + 2 + 3}} + ..... + \frac{2}{{k\left( {k + 1} \right)}}} \right) + \frac{2}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$$
$$ = \frac{{2k}}{{k + 1}} + \frac{2}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$$     [using equation (i)]
$$ = \frac{{2k\left( {k + 2} \right) + 2}}{{\left( {k + 1} \right)\left( {k + 2} \right)}} = \frac{{2\left[ {{k^2} + 2k + 1} \right]}}{{\left( {k + 1} \right)\left( {k + 2} \right)}}$$       [taking 2 common in numerator part]
$$ = \frac{{2{{\left( {k + 1} \right)}^2}}}{{\left( {k + 1} \right)\left( {k + 2} \right)}} = \frac{{2\left( {k + 1} \right)}}{{k + 2}} = \frac{{2\left( {k + 1} \right)}}{{\left( {k + 1} \right) + 1}}$$
Therefore, $$P\left( {k + 1} \right)$$   is true, when $$P\left( {k} \right)$$  is true.
Hence, from the principle of mathematical induction, the statement is true for all natural numbers $$n.$$

Since, $$P\left( 1 \right):2 < 1$$   is false; $$P\left( 2 \right):{2^2} < 1 \times 2$$    is false; $$P\left( 3 \right):{2^3} < 1 \times 2 \times 3$$     is false; $$P\left( 4 \right):{2^4} < 1 \times 2 \times 3 \times 4$$      is true.

Given that, $$P\left( n \right):{3^n} < n!$$    Now, $$P\left( 7 \right):{3^7} < 7!$$    is true
$$\eqalign{ & {\text{Let, }}P\left( k \right):{3^k} < k! \cr & \Rightarrow P\left( {k + 1} \right):{3^{k + 1}} \cr & = 3 \cdot {3^k} < 3 \cdot k! < \left( {k + 1} \right)!\,\,\left( {\because k + 1 > 3} \right) \cr} $$

$$\eqalign{ & {2^4} \equiv 1\left( {\bmod 5} \right) \cr & \Rightarrow {\left( {{2^4}} \right)^{75}} \equiv {\left( 1 \right)^{75}}\left( {\bmod 5} \right) \cr & {\text{i}}{\text{.e}}{\text{., }}{2^{300}} \equiv 1\left( {\bmod 5} \right) \cr & \Rightarrow {2^{300}} \times 2 \equiv \left( {1.2} \right)\left( {\bmod 5} \right) \cr & \Rightarrow {2^{301}} \equiv 2\left( {\bmod 5} \right) \cr} $$
$$\therefore $$ Least positive remainder is 2.

$${\text{Put, }}n = 1,{\left( {1 + \frac{1}{1}} \right)^1} = 2$$
$$\therefore {\left( {1 + \frac{1}{n}} \right)^n} > {2},n$$     is a positive integer.

Let $$P\left( n \right):{S_n}$$
\[ = \left\{ \begin{array}{l} \frac{{n{{\left( {n + 1} \right)}^2}}}{2},\,\,{\rm{when }}\,\,n\,\,{\rm{is\,\, even}}\\ \frac{{{n^2}\left( {n + 1} \right)}}{2},{\rm{ when }}\,\,n\,\,{\rm{is \,\,odd}} \end{array} \right.\]
Also, note that any term $$T_n$$ of the series is given by
\[{T_n} = \left\{ \begin{array}{l} {n^2},{\rm{ if }}\,\,n\,\,{\rm{ is \,\,odd}}\\ {\rm{2}}{{\rm{n}}^2},{\rm{ if }}\,\,n\,\,{\rm{ is\,\, even}} \end{array} \right.\]
We observe that $$P\left( 1 \right)$$  is true, Since
$$P\left( 1 \right):{S_1} = {1^2} = 1 = \frac{{1 \cdot 2}}{2} = \frac{{{1^2} \cdot \left( {1 + 1} \right)}}{2}$$
Assume that $$P\left( k \right)$$  is true for some natural number $$k,$$ i.e.,
Case I : When $$k$$ is odd, then $$k + 1$$  is even. We have,
$$\eqalign{ & P\left( {k + 1} \right):{S_{k + 1}} = {1^2} + 2 \times {2^2} + ..... + {k^2} + 2 \times {\left( {k + 1} \right)^2} \cr & = \frac{{{k^2}\left( {k + 1} \right)}}{2} + 2 \times {\left( {k + 1} \right)^2} \cr & = \frac{{\left( {k + 1} \right)}}{2}\left[ {{k^2} + 4\left( {k + 1} \right)} \right] = \frac{{k + 1}}{2}\left[ {{k^2} + 4k + 4} \right] \cr & = \frac{{k + 1}}{2}{\left( {k + 2} \right)^2} = \left( {k + 1} \right)\frac{{{{\left[ {\left( {k + 1} \right) + 1} \right]}^2}}}{2} \cr} $$
So, $$P\left( {k + 1} \right)$$   is true, whenever $$P\left( {k} \right)$$  is true, in the case when $$k$$ is odd.
Case II : When $$k$$ is even, then $$k + 1$$  is odd.
Now, $$P\left( {k + 1} \right):{S_{k + 1}} = {1^2} + 2 \times {2^2} + ..... + 2 \cdot {k^2} + {\left( {k + 1} \right)^2}$$
$$ = \frac{{k{{\left( {k + 1} \right)}^2}}}{2} + {\left( {k + 1} \right)^2} = \frac{{{{\left( {k + 1} \right)}^2}\left( {\left( {k + 1} \right) + 1} \right)}}{2}$$
Therefore, $$P\left( {k + 1} \right)$$   is true, whenever $$P\left( {k} \right)$$  is true for the case when $$k$$ is even.
Thus, $$P\left( {k + 1} \right)$$   is true whenever $$P\left( {k} \right)$$  is true for any natural number $$k.$$ Hence, $$P\left( {n} \right)$$  true for all natural numbers $$n.$$

Check through option, the condition $$3^n > n^3$$  is true when $$n \geqslant 4.$$

$$\eqalign{ & S\left( k \right) = 1 + 3 + 5 + ...... + \left( {2k - 1} \right) = 3 + {k^2} \cr & S\left( 1 \right):1 = 3 + 1,\,\,{\text{which is not true}} \cr & \because \,\,S\left( 1 \right)\,\,{\text{is not true}}{\text{.}} \cr & \therefore \,\,{\text{P}}{\text{.M}}{\text{.I}}{\text{. cannot be applied}} \cr & {\text{Let }}S\left( k \right){\text{ is true, i}}{\text{.e}}{\text{.,}} \cr & 1 + 3 + 5 + ...... + \left( {2k - 1} \right) = 3 + {k^2} \cr & \Rightarrow \,\,1 + 3 + 5 + ...... + \left( {2k - 1} \right) + 2k + 1 \cr & = 3 + {k^2} + 2k + 1 = 3 + {\left( {k + 1} \right)^2} \cr & \therefore \,\,S\left( k \right) \Rightarrow S\left( {k + 1} \right) \cr} $$