I want to prove the arithmetic series formula using the fact that its elements taken in pairs in a certain way all sum to the same number. The equation for each term of this sequence is an=a1+(n−1)dan=a1+(n-1)d. So for a1=3a1=3, d=2d=2 and n=5n=5 we get a series Sn=3+5+7+9+11Sn=3+5+7+9+11 By

**empirical **studies we notice that in any arithmetic series with

**even **number of elements we have n2n2 pairs of the same value, namely (an+a1)(an+a1). Hence with no doubt we can simply write, the entire sum is equal to n2⋅(an+a1)n2⋅(an+a1). Now what if nn is odd? Then we can say that still there is even numbers of pairs + one more term, so our sum for odd number of elements is n−12⋅(an+a1)+Xn-12⋅(an+a1)+X. Since we know that the formula for even number of elements is valid for odd number of elements as well (proof

@HallsofIvy showed) here w need to show that X=an+a12X=an+a12 so we get n−12⋅(an+a1)+an+a12=n2⋅(an+a1)n-12⋅(an+a1)+an+a12=n2⋅(an+a1). So first we need to find out which element is the remaining one and then, get its value. The element which remains unpaired is the one which has equal number of elements on its left and right side, we know it is element with index n+12n+12 but hmm can we prove it? What

@lev888 derived is almost that, once we prove XX is the middle term with index n+12n+12 we can find its value using

@lev888 formulas - its value will be the same as the average of first and last element. So the goal is to prove index of missing element. How to show that?