Consider An = (1+sqrt(2))^n + (1-sqrt(2))^n
Expanding both terms using the binomial theorem, notice that the odd powers cancel, while the coefficients of even powers are all integers, and therefore An is an integer.
Then, |1 -sqrt(2)| < 1, and so (1 -sqrt(2))^ n tends to zero as n tends to infinity.
Using logarithms and/or a calculator, we find that 10^-1149 < (1 -sqrt(2))^3000 < 10^-1148.
Therefore (1 + sqrt(2))^3000 has 1148 nines to the right of the decimal point, and so the 1000th such digit is a 9.
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