Expansion is tedious

The coefficient of $x^6$ will be the same for all $n \geq 6$ , so we can assume $n \to \infty$ . The expression becomes $\left(e^{x}\right)^{2} = e^{2x}$ .

The coefficient of $x^6$ in the Taylor expansion of $e^{2x}$ will be $\frac{2^{6}}{6!} = \frac{64}{720} = \frac{4}{45}$ .

$\sqrt{4 + 45} = \boxed{7}$ $_\square$

The $n$ th term in the expansion of the square is $A_n x^n = \sum_{k=0}^n \frac{x^k}{k!}\cdot\frac{x^{n-k}}{(n-k)!} = \sum_{k=0}^n \frac{x^n}{k!(n-k)!}.$ Divide out $x^n$ and multiply by $n!$ : $n! A_n = \sum_{k=0}^n \frac{n!}{k!(n-k)!} = \sum_{k=0}^n \left(\begin{array}{c} n \\ k \end{array}\right).$ In other words, we must sum over an entire row in Pascal's triangle. It is well known that the sum equal a power of two: $n! A_n = 2^n,$ from which it follows that $A_n = \frac{2^n}{n!}.$ In the case of $n = 6$ , $A_6 = \frac{2^6}{6!} = \frac{2^6}{(1\cdot 3\cdot 5)\cdot(2^3\cdot 3!)} = \frac{2^3}{3\cdot 5\cdot 2\cdot 3} = \frac{2^2}{3^2\cdot 5}.$ It follows that $a = 4, b = 45$ , and $\sqrt{a+b} = 7$ .

We know that the ${x}^{6}$ term in the expansion will be the sum of the products of all terms in the parentheses with orders that add up to 6. We can view the problem as taking the sum of the products of the coefficients of these terms. To account for the fact that we are squaring the sequence, we must multiply every term in our sum by 2 (except for the term obtained by multiplying the coefficients of terms with equal orders). The problem is now in the form $2(\frac{1}{6!} + \frac{1}{1!5!} + \frac{1}{2!4!)} + \frac{1}{3!3!})$ . This expression can be simplified to the fraction $\frac{4}{45}$ . $\sqrt{4 + 45} = \sqrt{49} = \boxed{7}$

Just take the limit as $n\rightarrow\infty$ of the whole expression and you get $e^{2x}$ and the coefficient of $x^6$ of the Taylor expansion of $e^{2x}$ is $4/45$ and taking square root of 49 gives you 7 and that's the answer

$Only~terms~that~give~x^6~are~given~below.\\ 1*\dfrac{x^6}{6!}~~+~~\dfrac{ x} {1!}*\dfrac{x^5}{5!}~~+~~\dfrac {x^2} {2!}*\dfrac{x^4}{4!}~~+~~\dfrac{ x^3}{3!}*\dfrac{x^3}{3!} ~~+~~\dfrac{ x^4} {4!}*\dfrac{x^2}{2!}~~+~~\dfrac {x^5} {5!}*\dfrac{x}{1!}~~+~~\dfrac{ x^6}{6!}*1\\ =2* \Big \{1*\dfrac{x^6}{6!}~~+~~\dfrac{ x} {1!}*\dfrac{x^5}{5!}~~+~~\dfrac {x^2} {2!}*\dfrac{x^4}{4!} \Big \}~~+~~\dfrac{ x^3}{3!}*\dfrac{x^3}{3!}\\ =x^6 \Big \{ \dfrac2 {6!} ~~+~~ \dfrac2 {5!}~~+~~\dfrac {2} {2!*4!}~~+~~\dfrac 1 { 3!*3!} \Big \}\\ =x^6 \Big \{ \dfrac1 {360} ~~+~~ \dfrac1 {60}~~+~~\dfrac 1 {24}~~+~~\dfrac 1 { 36} \Big \}\\ =\dfrac{x^6}{360}*\{1+6+15+10\}=x^6\dfrac4{45}= x^6\dfrac a b. \\ \therefore~\sqrt{4+45}=\large \color{#D61F06}{7}$

Write the above expansion as $\zeta$ =

$(e^{x} - \sum_{i+1}^\infty x^{n+i}/(n+i)!)^{2}$

= $e^{2x} - (terms- not -containing : x^{6})$

$\implies$ $coeff x^{6} in : \zeta$ = $64/6!$ = $4/45$

Therefore answer is $\sqrt{4+45}= \sqrt {49}= \boxed{7}$