Solution:

Since d₁ is not 0 and 5, d₂ is equal to 2, 4, 6, or 8. Then the sequence d₂, d₃, ... is periodic with period 4. Therefore, for each n>1

    a_n+4 = a_n +2+4+6+8

and a_n+4p =a_n + 20p for any natural number p. The sequence {a_n} includes a term divisible by 4, say a_n1 =4k. But then a_n1+4p = 4k +20p = 4(k+5p). Since 2^m (mod5) m=0, 1, 2, ... is a periodic sequence 1, 2, 4, 3, 1, 2, 4, 3, ..., infinitely many terms 2^m are equal to k mod(5). Therefore, k+5p contains infinitely many terms of the form 2^m.