How to deduce the summation formula of arithmetic sequence with complete square formula

How to deduce the summation formula of arithmetic sequence with complete square formula

A sequence (an), if from its second term, the difference between each term and its previous term is equal to the same constant, that is, an -- an-1 = D (n = 2, n belongs to n *). Where D is a constant, we call such a sequence an arithmetic sequence, and D is the tolerance of arithmetic sequence
After the expansion of the complete square formula, is there such a relationship between the items (satisfying the arithmetic sequence)? So I think what you are talking about should be the binomial theorem

What is the summation formula of arithmetic sequence

(first item + last item) * number of items / 2

The difficult problem of summation of arithmetic sequence
Let {an} be an arithmetic sequence, and prove that the sequence {BN} with BN = a1 + A2 +... An / N (n belongs to N +) as the general term formula is an arithmetic sequence

a(n)=a+(n-1)d,
b(n)=[a(1)+...+a(n)]/n = [na+n(n-1)d/2]/n = a+ (n-1)(d/2).
The first term of {B (n)} is a (1) = a, and the tolerance is D / 2