You need to show that for

*any* term b in B and

*any* term c in C there exists a term e in E such that b + c = e. So, it is not enough to consider the situation when both b and c are the nth terms, i.e., when b = 2 + 5(n - 1) and c = 3 + 5(n - 1) for the same n.

An arbitrary term b in B has the form 2 + 5(m - 1) for some m ≥ 1 and an arbitrary term c in C has the form 3 + 5(n - 1) for some n ≥ 1. Then b + c = 5 + 5(m + n - 2). To prove that b + c is in E, we need to find some k ≥ 1 such that b + c = 5 + 5(k - 1), so we need 5 + 5(m + n - 2) = 5 + 5(k - 1). This equation is equivalent to k = m + n - 1. Since m + n - 1 ≥ 1, we found the required k.

See

this post and

Wikibooks. In plain text, it is acceptable to write B_n for

and x^2 for

.