Solution to puzzle 27: 1000th digit

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 a_n 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.

Remarks

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.

Additional puzzle

Find the first digit before and after the decimal point in ( + )³⁰⁰⁰.

Source: Inspired by Binet's formula for Lucas numbers