Beware of Surds

Relevant wiki .- Fibonacci sequence

This is the $11^{th}$ term in Fibonacci sequence. $F_1 = F_2 = 1, \space F_{n +1} = F_{n} + F_{n -1}$ . This is a difference equation... $x^2 -x - 1 = 0 ....\Rightarrow F_n = \frac{(\frac{1 + \sqrt{5}}{2})^{n} - (\frac{1 - \sqrt{5}}{2})^{n}}{\sqrt{5}} \Rightarrow F_{11} = \frac{(\frac{1 + \sqrt{5}}{2})^{11} - (\frac{1 - \sqrt{5}}{2})^{11}}{\sqrt{5}} = 89$

$\text{Solve it Normally} \\ \text{Let A} = \dfrac{\left(\dfrac{1+\sqrt5}{2}\right)^{11}-\left(\dfrac{1+\sqrt5}{2}\right)^{11}}{\sqrt{5}} \\ A = \dfrac{1}{2^{10}} \left( \dfrac{ \dbinom{11}{1} \sqrt5 +\dbinom{11}{3} 5^{\frac32}+\dbinom{11}{5} 5^{\frac52}+\cdots+\dbinom{11}{11} 5^{\frac{11}{2}}}{\sqrt5} \right) \\ A = \dfrac{1}{2^{10}} \left[ \dbinom{11}{1} +\dbinom{11}{3} 5+\dbinom{11}{5} 25+\cdots+\dbinom{11}{11} 5^5 \right] \\ A=\dfrac{1}{2^{10}}\left[ (11+3125)+\dfrac{11\times 10\times9}{3!}\times5+\dfrac{11\times 10\times 9 \times 8\times 7}{5!}\times25+\dfrac{11\times 10 \times 9\times 8}{4!}\times125+\dfrac{11\times10}{2}\times625 \right] \\ A= \dfrac{1}{2^{10}} \left[ 3136+\underbrace{11\times 25\times 3+11\times6 \times 7\times 25+330\times 125+625\times55}_{11\times5\times5\left( 3+42+150+125\right)} \right] \\ A=\dfrac{1}{2^{10}} \left[3136+11\times25\times320 \right] \implies \dfrac{1}{2^4}\left(49+11\times25\times5 \right) = \boxed{89}$

$\text{This method Requires more steps but it is less time consuming if you do it the right way}$