So recently I had figured out on my own that: $$1 + 2 + \cdots + n = P \iff 2^{n - 1}(2^n - 1) = \{P : P = \text{Perfect Number}\}$$ Now I had figured out something else as well: $$1 - 2^2 + \cdots - (2^n - 2)^2 + (2^n - 1)^2 = P \iff 2^{n - 1}(2^n - 1) = P$$ I have made an attempt of trying to see why this works, and I made a formula from it, which is in the yellow box along with the question underneath (if you want to skip my approach of how I discovered the formula).
My Approach: $$1 - 2^2 +\cdots - (2^n - 2)^2 + (2^n - 1)^2 = P \iff 1 + 2 + \cdots + n = P \ \land \ 2^{n - 1}(2^n - 1) = P$$ $$\begin{align} \therefore P &= (1 + 2)(1 - 2) + (3 + 4)(3 - 4) + \cdots + (2^n - 2 + 2^n - 1)\big(2^n - 2 - (2^n - 1)\big) \\ &= (-1)(1 + 2) + (-1)(3 + 4) + \cdots + (-3)(2^{n + 1} - 3)(\require{cancel}\cancel{2^n} - 2 \cancel{- 2^n} + 1) \\ &= -(1 + 2) - (3 + 4) - \cdots + (-1)(2^{n + 1} - 3) \\ &= -3 - 7 - \cdots - (2^{n + 1} - 3) \\ &= \cancel{-3} - 7 - \cdots - 2^{n + 1} \cancel{+ 3} \\ &= -7 - \cdots - 2^{n + 1} \\ &< 0 \end{align}$$ Obviously I did something wrong, so I looked back at it and tried again. Consider the following alternating sum: $$1 - 2^2 + 3^2 - 4^2 + 5^2 - 6^2 + 7^2 - 8^2 +\cdots$$ Let's look at every odd term and every even term: $$\overbrace{+ 1}^{1} \ \overbrace{- 2^2}^{2} \ \overbrace{+ 3^2}^{3} \ \overbrace{- 4^2}^{4} \ \overbrace{+ 5^2}^{5} \ \overbrace{- 6^2}^{6} \ \overbrace{+ 7^2}^{7} \ \overbrace{- 8^2}^{8} +\cdots$$ Every odd term makes the expression greater than $0$, and every even term makes the expression lesser than $0$. Since $P > 0$, then the expression must end on an odd term, being $(2^n - 1)^2$. This then made me realise that $n$ had to be a prime number because in the expression $2^n - 1$, if $n$ is not prime, then $n = ab$ and thus $(2^n - 1)$ is not prime. $$\begin{align} 2^{ab} - 1 = (2^a)^b - 1 &= (2^a - 1)\big((2^a)^{b - 1} + (2^a)^{b - 2} + \cdots + (2^a)^{b - (b - 1)} + 2^a + 1\big) \\ &= (2^b - 1)\big((2^b)^{a - 1} + (2^b)^{a - 2} + \cdots + (2^b)^{a - (a - 1)} + 2^b + 1\big) \\ \text{However}, \ 2^n - 1 &= \{M_p : M_p = \text{Mersenne Prime}, \ \because P = 2^{n - 1}(2^n - 1)\} \end{align}$$. $\qquad\qquad\qquad\qquad\qquad \ \ \longrightarrow$ Mersenne Prime - Wikipedia
$$\begin{align} \therefore P &= \{-3 - 7 - 11 - \cdots + (2^n - 1)^2 : P = 2^{n - 1}(2^n - 1) = \text{Perfect Number}\} \\ &= (2^n - 1)^2 + \sum_{k = 1}^{(2^n - 2)/2}(1 - 4k) \\ &= (2^n - 1)^2 - \sum_{k = 1}^{2^{n - 1} - 1}(4k - 1) \\ &> 0 \end{align}$$ Did I just discover a new formula for a perfect number (unless this formula has been discovered by someone else, which I believe it has)? Why do perfect numbers hold so many fascinating properties?
If you are using any other mathematical techniques to answer my question that are seperate from this post, they may be foreign to my understanding of mathematics so far, so bearing that in mind, please be clear and try not to skip too many steps in the process.
Thank you in advance.
It only works when 2$^n$-1 is a mersenne prime.
e.g.n=4 it is not a perfect number.