Find the nth cross-alternate sum

Jelly, 21 19 11 10 7 bytes

²~³¡H+2

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Idea

Assume for a second that the first term of the final sum is subtracted rather than added.

Let n be a positive integer.

Even case

 1              6
    8       11
      15 16
      21 22
   26       29
31             36

The differences between the diagonal elements on the lower half of the rows are the first n ÷ 2 odd natural numbers. Since 1 + 3 + 5 + … + (2k + 1) = k², they sum to (n ÷ 2)² = n² ÷ 4.

In this example

- 1 + 6 - 8 + 11 - 15 + 16 - 21 + 22 - 26 + 29 - 31 + 36  =
-(1 - 6)-(8 - 11)-(15 - 16)-(21 - 22)-(26 - 29)-(31 - 36) =
(    5   +   3    +    1  )+(   1    +    3    +    5   ) =
             9             +              9               = 18

Thus, the sum is 2 × n² ÷ 4 = n² ÷ 2.

Odd case

 1           5
    7     9
      13
   17    19
21          25

The differences between the diagonal elements on the corresponding rows from above and below (1 and 5, and 21 and 25; 7 and 9, and 17 and 19) is the same, so they will cancel out in the alternating sum.

In this example

- 1 + 5 - 7 + 9 - 13 + 17 - 19 + 21 - 25  =
-(1 - 5)-(7 - 9)- 13 +(17 - 19)+(21 - 25) =
    4   +   2   - 13 -    2    -    4     = -13

All that's left is the negative of the central element, which is the arithmetic mean of the first and last number, so it can be calculated as -(n² + 1) ÷ 2.

General case

Since ~x = -(x + 1) for two's complement integers (~ denotes bitwise NOT), the formula for the odd case can be rewritten as ~n² ÷ 2.

Also, since the first term (1) of the original sum is added instead of subtracted, the above formulas leave an error of 2, which has to be corrected.

Therefore, the n^th cross-alternate sum is n² ÷ 2 + 2 if n is even, and ~n² ÷ 2 + 2 if it is odd.

Finally, bitwise NOT is an involution, i.e., ~~x = x for all x. This way ~~~x = ~x, ~~~~x = x, and, in general, ~ⁿx (meaning that ~ is applied n times) is x if n is even and ~x if it is odd.

Thus, we can rewrite our general formula as ~ⁿn² ÷ 2 + 2 for all positive integers n.

Code

²~³¡H+2    Main link. Input: n

²          Yield n².
 ~         Apply bitwise NOT to n²...
  ³¡           n times.
    H      Halve the result.
     +2    Add 2.

JavaScript, 40 38 22 bytes

Using that new-fangled, fancy closed form solution that's all the rage!

n=>(n%2?3-n*n:4+n*n)/2

Thanks to ThomasKwa, I can eliminate my costly recursive function.

Jelly, 12 bytes

RṖµ²+‘×-*$SC

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