Given $$S(K)$$ $$ = 1 + 3 + 5... + \left( {2K - 1} \right) = 3 + {K^2}$$

When k = 1, S(1): 1 = 3 + 1,

L.H.S of S(k) $$ \ne $$ R.H.S of S(k)

So S(1) is not true.

As S(1) is not true so principle of mathematical induction can not be used.

S(K+1) = 1 + 3 + 5... + (2K - 1) + (2K + 1) = 3 + (k + 1)²

Now let S(k) is true

$$\therefore$$ 1 + 3 + 5 +........(2k - 1) = 3 + k²

$$ \Rightarrow $$ 1 + 3 + 5 +........(2k - 1) + (2k + 1) = 3 + k² + 2k +1

= 3 + (k + 1)²

$$ \Rightarrow $$ S(k + 1) is true.

$$\therefore$$ S(k) $$ \Rightarrow $$ S(k + 1)

General term = $${}^{256}{C_r}.{\left( {\sqrt 3 } \right)^{256 - r}}.{\left( {\root 8 \of 5 } \right)^r}$$
= $${}^{256}{C_r}.{\left( 3 \right)^{{{256 - r} \over 2}}}.{\left( 5 \right)^{{r \over 8}}}$$

When $${{{256 - r} \over 2}}$$ is integer then $${\left( 3 \right)^{{{256 - r} \over 2}}}$$ is integer.
And when $${{r \over 8}}$$ is integer then $${\left( 5 \right)^{{r \over 8}}}$$ is integer.

Entire general term will be integer when $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ both are integer.

$${{{256 - r} \over 2}}$$ is integer when r = 0, 2, 4, 6, ......, 256

$${{r \over 8}}$$ is integer when r = 0, 8, 16 ,......., 256

Now both $${{{256 - r} \over 2}}$$ and $${{r \over 8}}$$ will be integer when r = 0, 8, 16, ...., 256 (This is an AP)

$$\therefore$$ 256 = 0 + (n - 1)8 using formula of AP, t_n = a + (n - 1)d

$$\therefore$$ n = $${{256} \over 8} + 1$$ = 32 + 1 = 33

General term of $${\left( {1 + x} \right)^{n}}$$ is ($${T_{r + 1}}$$) = $${{n\left( {n - 1} \right).....\left( {n - r + 1} \right)} \over {1.2.3....r}}{x^r}$$

$$\therefore$$ General term of $${\left( {1 + x} \right)^{27/5}}$$ = $${{{{27} \over 5}\left( {{{27} \over 5} - 1} \right).....\left( {{{27} \over 5} - r + 1} \right)} \over {1.2.3....r}}{x^r}$$

For first negative term, $${\left( {{{27} \over 5} - r + 1} \right)}$$ < 0

$$ \Rightarrow r > {{27} \over 5} + 1$$

$$ \Rightarrow r > {{32} \over 5}$$

$$ \Rightarrow r > 6.4$$

$$\therefore$$ r = 7

$${T_{7 + 1}} = {T_8}$$ means 8^th term is the first negative term.

Given $${a_n} = \sqrt {7 + \sqrt {7 + \sqrt {7 + .......} } } $$

$$\therefore$$ $${a_n} = \sqrt {7 + {a_n}} $$

$$ \Rightarrow $$ $$a_n^2 = 7 + {a_n}$$

$$ \Rightarrow $$ $$a_n^2 - {a_n} - 7 = 0$$

$$ \Rightarrow {a_n} = {{1 \pm \sqrt {1 - 4 \times 1 \times - 7} } \over 2}$$

$$ \Rightarrow {a_n} = {{1 \pm \sqrt {29} } \over 2}$$

As $${a_n}$$ > 0,

$$\therefore$$ $${a_n} = {{1 + \sqrt {29} } \over 2}$$ = 3.19

So $${a_n} > 3\,\,\forall \,\,n \ge 1$$