Question

Without a calculator how would you solve this?

Answer 1

Answer:

the largest number that is less than (2+√3)^6 is 2701

Step-by-step explanation:

$\bigstar \ (2+\sqrt{3})^6= \\\\C{}^{0}_{6}2^{6}+C{}^{1}_{6}2^{5}\sqrt{3}^1+C{}^{2}_{6}2^{4}\sqrt{3}^2+C{}^{3}_{6}2^{3}\sqrt{3}^3+C{}^{4}_{6}2^{2}\sqrt{3}^4+C{}^{5}_{6}2^{1}\sqrt{3}^5+C{}^{6}_{6}\sqrt{3}^6$

$=64+192\sqrt{3}+720 +480\sqrt{3} +540+108\sqrt{3} +27\\=1351+780\sqrt{3}$

$\bigstar\ (2-\sqrt{3} )^6 =\\\\1351-780\sqrt{3}$

$\bigstar\ \bigstar\ (2+\sqrt{3} )^6+(2-\sqrt{3} )^6 =\\\\1351+780\sqrt{3}+1351-780\sqrt{3}$

$\Longrightarrow(2+\sqrt{3} )^6+(2-\sqrt{3} )^6 =2702$

$\Longrightarrow(2+\sqrt{3} )^6 =2702-(2-\sqrt{3} )^6$

$0 < \left( 2-\sqrt{3} \right) < 1 \Longrightarrow 0 < \left( 2-\sqrt{3} \right)^{6} < 1 \Longrightarrow-1 < \left( 2-\sqrt{3} \right)^{6} < 0$

$\Longrightarrow2702-1 < 2702-\left( 2-\sqrt{3} \right)^{6} < 2702+0$

$\Longrightarrow 2701 < \left( 2+\sqrt{3} \right)^{6} < 2702$