I looked at formulas concerning prime numbers, and I came across Mersenne Primes which are primes of the form $2^n - 1 : n\in \mathbb{N}$ and the symbol used to denote a mersenne prime is $M_p$.
Along with mersenne primes, I came across the formula: $$(2^n - 1)(2^{n - 1}) = \{P : n \in \mathbb{N}, \ 2^n - 1 = M_p \ \land \ P =\text{Perfect Number}\}$$ but I also found out something else which I didn't come across, and I don't want to say that I discovered it because I am sure someone else must have discovered it: $$1 + 2 +\cdots+x = P \iff 2^n - 1 = x = \{M_p : n\in \mathbb{N}\}$$ For example; $$1 + 2 + 3 = 6 \implies 6\text{ is a perfect number}\because 2^2 - 1 = 3 = \text{Mersenne Prime}$$ $$1 + 2 + 3 + 4 + 5 + 6 + 7 = 28 \implies 28\text{ is perfect}\because 2^3 - 1 = 7 = \text{Mersenne Prime}$$ However, every time I try to research something like this, this symbol is always related to it $\longrightarrow \sum$ but I do not know what it means, so I have made an attempt of trying to see if it works.
My Attempt: $$\begin{align} P &= 1 + 2 + \cdots + x \\ &= 1 + 2 + \cdots + 2^n - 1 \\ &= 2 + \cdots + 2^n \\ &= 2(2^{n - 1} + 1) + \cdots \end{align}$$ I can actually get every even number in the sum and put it in the bracket, and every odd number and put that outside the bracket: $$\implies 2(2^{n - 1} + 1 + 2 + \cdots + x_1) + 3 + 5 + \cdots + x_2 = P : x_1 \in \mathbb{N} \ \land \ \color\purple{x_2 \in \mathbb{N}_{2n - 1, \ \forall n \in \mathbb{N}}}$$. $$\because \overbrace{o_1 + o_2 + \cdots + o_n}^{\text{$n$ times}} = n^2 : o_n = \text{$n^{th}$ odd number}$$ $$\begin{align} \implies 2(2^{n - 1} + 1 + 2 + \cdots + x_1) + (x_2^{\text{ $ \ 2$}} - 1) &= P \because o_1 = 1 \\ \implies 2(2^{n - 1} + 1 + 2 + \cdots + x_1) &= P - (x_2^{\text{ $ \ 2$}} - 1) \\ &= P - x_2^{\text{ $ \ 2$}} + 1 \\ \implies 2^{n - 1} + 1 + 2 + \cdots + x_1 &= \frac{P - x_2^{\text{ $ \ 2$}} + 1}{2} \\ &= \frac{P}{2} - \frac{x_2^{\text{ $ \ 2$}}}{2} + \frac{1}{2} \\ \implies 2^{n - 1} + \frac{1}{2} + 2 + 3 + \cdots + x_1 &= \frac{P}{2} - \frac{x_2^{\text{ $ \ 2$}}}{2} \\ \implies \frac{2^n + 1}{2} + 2 + 3 + \cdots + x_1 &= \frac{P - x_2^{\text{ $ \ 2$}}}{2} \\ \implies 2^n + 1 + 4 + 6 + \cdots + 2x_1 &= P - x_2^{\text{ $ \ 2$}} \\ \implies 2^n + 1 + 2(2 + 4 + \cdots + x_1) &= P - x_2^{\text{ $ \ 2$}} \\ \implies 2(2^{n - 1} + 2 + 4 + \cdots + x_1) + 1 &= P - x_2^{\text{ $ \ 2$}} \\ \implies 2^{n - 1} + 2 + 4 + \cdots + x_1 &= \frac{P - x_2^{\text{ $ \ 2$}} - 1}{2} \\ \implies 2(2^{n - 2} + 1 + 2 + \cdots + x_1) &= \frac{P - x_2^{\text{ $ \ 2$}} - 1}{2} \\ \implies 2(2^{n - 2} + P) &= \frac{P - x_2^{\text{ $ \ 2$}} - 1 }{2} \\ \implies 2^n + P &= P - x_2^{\text{ $ \ 2$}} - 1 \\ \therefore P &= P - x_2^{\text{ $ \ 2$}} - 1 - 2^n \\ &= P - (x_2^{\text{ $ \ 2$}} + 1 + 2^n) \\ \therefore \color\red{x_2^{\text{ $ \ 2$}} + 1 + 2^n} &\color\red{=} \color\red{0} \end{align}$$ Could you please explain to me where I made a mistake? I must have made it somewhere because the last equation highlighted in red doesn't make any sense! Surely this must be a contradiction unless their is a mistake. Any help would be appreciated, but I would mostly appreciate if you reveal the meaning of this symbol $\longrightarrow \sum$. I am in Year 9, so I may not be ready to know what it means... It looks Greek to me, but I don't know what letter it is in the Greek alphabet.
Also, did I write the part highlighted in purple correctly?
Thank you in advance.