Question

find b

Answer 1

\[(2+\sqrt{3})^n=5042+b \sqrt{3}\]

Answer 2

think of it as a serial
Do you think? @mukushla

Answer 3

yes

Answer 4

\[n \in \mathbb{Z}^+ \] like this?

Answer 5

yes

Answer 6

think of \[(2-\sqrt{3})^n\]

Answer 7

\[(2+\sqrt{3})^n=5042+b \sqrt{3} \\ then \ \ (2-\sqrt{3})^n=5042-b \sqrt{3} \\ so \ \ ((2+\sqrt{3})(2-\sqrt{3}))^n=(5042+b \sqrt{3})(5042-b \sqrt{3}) \\ then \ \ \ (5042+b \sqrt{3})(5042-b \sqrt{3})=1 \\ 5042^2-3b^2=1 \\ b=2911\] im not sure!!!