Determine the 521st digit to the right of the decimal point in the base ten representation of:
(1+√2)2022
(sqrt(2)+1)^2022 + (sqrt(2)-1)^2022 is an integer.
To see this expand using the binomial theorem.
Notice that all the negative terms in the expansion of (sqrt(2)-1)^2022 are the terms where sqrt(2) is raised to an odd power.
Then all the terms with radicals cancel out with the corresponding terns from (sqrt(2)+1)^2022
sqrt(2)-1 is between 0 and 1, so its powers will also be in that range.
(sqrt(2)-1)^3 ~= 0.071, importantly this is less than 0.1
Then 0 < (sqrt(2)-1)^2022 < 0.1^674.
Therefore at least the first 674 digits after the decimal point of (sqrt(2)-1)^2022 are 0s.
Then since (sqrt(2)+1)^2022 + (sqrt(2)-1)^2022 is an integer, subtracting the (sqrt(2)-1)^2022 back out leaves a number with at least 674 9's after the decimal point of (sqrt(2)+1)^2022. The 521st digit is one of those 9's.
Edited on July 28, 2023, 10:02 am