What is the 1000th digit to the right of the decimal point in the decimal representation of (1 + )3000 ?
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 − | < 1, and so (1 − )n tends to zero as n tends to infinity.
Using logarithms and/or a calculator, we find that 10−1149 < (1 − )3000 < 10−1148.
Therefore (1 + )3000 has 1148 nines to the right of the decimal point, and so the 1000th such digit is a 9.
Note that a large odd exponent would generate a string of zeroes rather than nines.
As a generalization, note that (a + b)n + (a − b)n is an integer for any positive integers a, b, and r. (Ignore the trivial case where r is a perfect square.)
Therefore as n tends to infinity, (a + b)n will tend to an integer if |a − b| < 1.
Find the first digit before and after the decimal point in ( + )3000.
Source: Inspired by Binet's formula for Lucas numbers