Solution:

Since d_{1} is not 0 and 5, d_{2}
is equal to 2, 4, 6, or 8. Then the sequence d_{2},
d_{3}, ... 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}.