We know, $$\,{\left( {a + b} \right)^n}$$ = $${}^n{C_0}.{a^n} + {}^n{C_1}.{a^{n - 1}}.b + ... + {}^n{C_n}.{b^n}$$

Remember to find sum of coefficient of binomial expansion we ave to put 1 in place of all the variable.

So put $$a$$ = b = 1

$$\therefore$$ 2ⁿ = $${}^n{C_0} + {}^n{C_1} + {}^n{C_2}... + {}^n{C_n}$$

According to question, 2ⁿ = 4096 = 2¹²

$$ \Rightarrow n = 12$$

So $$\,{\left( {a + b} \right)^n}$$ = $$\,{\left( {a + b} \right)^{12}}$$

Here n = 12 is even so formula for greatest term is
$${T_{{n \over 2} + 1}} = {}^n{C_{{n \over 2}}}.{a^{{n \over 2}}}.{b^{{n \over 2}}}$$

For n = 12, greatest term $${T_{6 + 1}} = {}^{12}{C_6}.{a^6}.{b^6}$$

$$\therefore$$ Coefficient of the greatest term = $${}^{12}{C_6}$$ = $${{12!} \over {6!6!}}$$ = 924

$${\left( {1 + 0.0001} \right)^{10000}}$$ = $${\left( {1 + {1 \over {{{10}^4}}}} \right)^{10000}}$$

= 1 + 10000$${ \times {1 \over {{{10}^4}}}}$$ + $${{10000\left( {9999} \right)} \over {2!}} \times {\left( {{1 \over {{{10}^4}}}} \right)^2}$$+......$$\infty $$

< 1 + 1 + $${1 \over {2!}}$$ + $${1 \over {3!}}$$ + ...... $$\infty $$ = e = 2.71828 < 3

$$\,{\left( {r + 2} \right)^{th}}$$ term = $${}^{2n}{C_{r+1}}{\left( x \right)^r}$$

And coefficient of $$\,{\left( {r + 2} \right)^{th}}$$ = $${}^{2n}{C_{r+1}}$$

$$3{r^{th}}$$ term = $${}^{2n}{C_{3r - 1}}{\left( x \right)^{3r - 1}}$$

And coefficient of $$3{r^{th}}$$ term = $${}^{2n}{C_{3r - 1}}$$

According to the question,
$${}^{2n}{C_{r+1}}$$ = $${}^{2n}{C_{3r - 1}}$$

$$ \Rightarrow \left( {r + 1} \right) + \left( {3r - 1} \right) = 2n$$

[As if $${}^n{C_p} = {}^n{C_q}$$ then p + q = n]

$$ \Rightarrow 4r = 2n$$

$$ \Rightarrow n = 2r$$

Here in this expansion $${\left( {1 + x} \right)^{p + q}}$$

The general term = $${T_{r + 1}} = {}^{p + q}{C_r}.{\left( x \right)^r}$$

$$\therefore$$ $${x^p}$$ will be present in the term = $${}^{p + q}{C_p}.{\left( x \right)^p}$$

So coefficient of $${x^p}$$ = $${}^{p + q}{C_p}$$

And $${x^q}$$ will be present in the term = $${}^{p + q}{C_q}.{\left( x \right)^q}$$

$$\therefore$$ coefficient of $${x^q}$$ = $${}^{p + q}{C_q}$$

We know $${}^n{C_r}$$ = $${}^n{C_{n - r}}$$

$$\therefore$$ $${}^{p + q}{C_q}$$ = $${}^{p + q}{C_{\left( {p + q} \right) - q}}$$ = $${}^{p + q}{C_p}$$

So coefficients of $${x^p}$$ and $${x^q}$$ are equal.